Table of Contents

QUADRATIC EQUATION
 Theory .............................................................................................................................................. 2

 Solved examples .............................................................................................................................. 6

 Exercise - 1 : Basic Objective Questions ......................................................................................... 12

 Exercise - 2 : Previous Year JEE Mains Questions .......................................................................... 18

 Exercise - 3 : Advanced Objective Questions ................................................................................ 21

 Exercise - 4 : Previous Year JEE Advanced Questions ................................................................... 28

 Answer Key ....................................................................................................................................... 33

 QUADRATIC EQUATION 2

QUADRATIC EQUATION

1. QUADRATIC EXPRESSION 3. NATURE OF ROOTS

The general form of a quadratic expression in x is,

(a) Consider the quadratic equation ax2 + bx + c = 0 where
2
f (x) = ax + bx + c, where a, b, c  R & a  0. a, b, c  R & a  0 then;
and general form of a quadratic equation in x is,
(i) D > 0  roots are real & distinct (unequal).
2
ax + bx + c = 0, where a, b, c  R & a  0.
(ii) D = 0  roots are real & coincident (equal).
2. ROOTS OF QUADRATIC EQUATION
(iii) D < 0  roots are imaginary..
(a) The solution of the quadratic equation,
(iv) If p + i q is one root of a quadratic equation,
then the other must be the conjugate p – i q &
2 b ± b2  4ac
ax + bx + c = 0 is given by x =
2a 
vice versa. p, q  R & i = -1 . 
The expression D = b 2 – 4ac is called the discriminant
of the quadratic equation. (b) Consider the quadratic equation ax2 + bx + c = 0 where
a, b, c  Q & a  0 then;
(b) If  &  are the roots of the quadratic equation
(i) If D > 0 & is a perfect square, then roots are
ax 2 + bx + c = 0, then ;
rational & unequal.
(i)  +  = – b/a (ii) α β = c/a
(ii) If α = p + q is one root in this case, (where p is
D
(iii) | α  β |= .
|a|
rational & q is a surd) then the other root must
(c) A quadratic equation whose roots are  &  is
(x –  ) (x –  ) = 0 i.e. be the conjugate of it i.e. β = p - q & vice versa.

x 2 – ( +  ) x +  = 0 i.e.

x 2 – (sum of roots) x + product of roots = 0.

Remember that a quadratic equation cannot have three

different roots & if it has, it becomes an identity.
y  (ax 2  bx  c)  a(x  ) (x  )

2
 b  D
 a x   
 2a  4a
QUADRATIC EQUATION 3

4. GRAPH OF QUADRATIC EXPRESSION 6. MAX. & MIN. VALUE OF QUADRATIC EXPRESSION

Consider the quadratic expression, y = ax 2 + bx + c, Maximum & Minimum Value of y = ax2 + bx + c occurs
at x = –(b/2a) according as :
a  0 & a, b, c  R then ;
For a > 0, we have :
(i) The graph between x, y is always a parabola.
If a > 0 then the shape of the parabola is
concave upwards & if a < 0 then the shape of
the parabola is concave downwards.

(ii) y > 0  x  R, only if a>0&D<0

 4ac - b 2 
(iii) y < 0  x  R, only if a<0&D<0 y ,  
 4a 

5. SOLUTION OF QUADRATIC INEQUALITIES

ax 2 + bx + c > 0  a  0  .

(i) If D > 0, then the equation ax 2 + bx + c = 0 has D b

y min  at x  , and y max  
4a 2a
two different roots (x 1 < x2).

Then a > 0  x (–, x1) (x2 , )

For a < 0, we have :
a<0  x (x1, x2)

a>0

x1 x2
x x
x1 x2
a<0  4ac  b 2 
y   , 
 4a 

P x
(ii) Inequalities of the form 0 can be
Qx
D b
y max  at x  , and y min   
4a 2a
quickly solved using the method of intervals
(wavy curve).
 4 QUADRATIC EQUATION

7. THEORY OF EQUATIONS

If 1, 2, 3, ....., n are the roots of the nth degree
polynomial equation : Remainder Theorem : If f (x) is a polynomial, then
f (h) is the remainder when f (x) is divided by x – h.
f (x) = a0xn + a1xn–1 + a2xn–2 + ...... + an–1x + an = 0
Factor theorem : If x = h is a root of equation
f (x) = 0, then x–h is a factor of f (x) and conversely.
where a 0, a 1, ....... a n are all real & a0  0,

Then,

a 9. MAX. & MIN. VALUES OF RATIONAL EXPRESSION

 α1 = – a 1 ;
0
Here we shall find the values attained by a rational
a
 α1 α 2 = + a 2 ; a1x 2  b1x  c1
0
expresion of the form for real values
a 2 x 2  b2 x  c2

a
 α1 α 2 α3 = – a 3 ; of x.
0
Example No. 4 will make the method clear.
............
10. COMMON ROOTS
a n
α1 α 2 α3 .....α n = (–1) n
a0 (a) Only One Common Root

Let  be the common root of ax 2 + bx + c = 0 &

8. LOCATION OF ROOTS
a’x 2 + b’x + c’ = 0, such that a, a’  0 and a b’  a’b.
Let f (x) = ax2 + bx + c, where a > 0 & a, b, c  R.
Then, the condition for one common root is :
(i) Conditions for both the roots of f(x) = 0 to be
greater than a specified number ‘k’ are : (ca  ca) 2 = (ab  a b) (bc  bc).
D 0 & f (k) > 0 & (–b/2a) > k.
(ii) Conditions for both roots of f (x) = 0 to lie on (b) Two Common Roots
either side of the number ‘k’ (in other words the Let  ,  be the two common roots of
numb e r ‘k’ l ies b etween the r o o ts o f
f (x) = 0 is: ax 2 + bx + c = 0 & a’x 2 + b’x + c’ = 0,
a f (k) < 0. such that a, a’  0.
(iii) Conditions for exactly one root of f (x) = 0 to lie
Then, the condition for two common roots is :
in the interval (k1, k2) i.e. k1 < x < k2 are :
D>0 & f (k1) . f (k2) < 0. a b c
 
a ' b ' c'
(iv) Conditions that both roots of f(x) = 0 to be
confined between the numbers k 1 & k 2 are
(k1 < k2) :
D  0 & f (k1) > 0 & f (k2) > 0 & k1 < (–b/2a) < k2.
QUADRATIC
5 EQUATION 5
QUADRATIC EQUATION

(iii) If there be any two real numbers ‘a’ & ‘b’ such
11. RESOLUTION INTO TWO LINEAR FACTORS
that f (a) & f (b) are of opposite signs, then
The condition that a quadratic function f (x) = 0 must have atleast one real root between
‘a’ and ‘b’.
f (x, y) = ax 2 + 2 hxy + by2 + 2 gx + 2 fy + c
(iv) Every equation f(x) = 0 of degree odd has
may be resolved into two linear factors is that ;
atleast one real root of a sign opposite to that
abc + 2fgh – af2 – bg2 – ch2 = 0 of its last term.
a h g 13. TRANSFORMATION OF EQUATIONS
OR h b f =0
g f c (i) To obtain an equation whose roots are reciprocals
of the roots of a given equation, it is obtained by
replacing x by 1/x in the given equation
12. FORMATION OF A POLYNOMIAL EQUATION (ii) Transformation of an equation to another
equation whose roots are negative of the roots
If 1, 2, 3, ....., n are the roots of the nth degree of a given equation–replace x by – x.
polynomial equation, then the equation is (iii) Transformation of an equation to another
n
x – S1x n–1
+ S2x n–2
+ S 3x n–3 n
+ ...... + (–1) Sn = 0 equation whose roots are square of the roots of a

where Sk denotes the sum of the products of roots given equation–replace x by x.

taken k at a time. (iv) Transformation of an equation to another
equation whose roots are cubes of the roots of a
Particular Cases
given equation–replace x by x1/3.
(a) Quadratic Equation if ,  be the roots the
quadratic equation, then the equation is :

x2 – S1x + S2 = 0 i.e. x2 – ( + ) x +  = 0

(b) Cubic Equation if , ,  be the roots the cubic

equation, then the equation is :

x3 – S1x2 + S2x – S3 = 0 i.e.

x3 – ( + +) x2 + ( +  + ) x –  = 0

(i) If  is a root of equation f(x) = 0, the polynomial

f (x) is exactly divisible by (x – ). In other words,
(x – ) is a factor of f(x) and conversely.

(ii) Every equation of nth degree (n  1) has exactly

n roots & if the equation has more than n roots,
it is an identity.
 QUADRATIC
6 EQUATION 6
QUADRATIC EQUATION

SOLVED EXAMPLES

Example – 1 Example – 2

If the remainder on dividing x3 + 2x2 + kx + 3 by x – 3 is 21, Find all the zeros of the polynomial x4 + x3 – 9x2 –3x + 18 if
find the quotient and the value of k. Hence find the zeros
it is given that two of its zeros are  3 and 3.
of the cubic polynomial x3 + 2x2 + kx – 18.
4 3 2
3 2 Sol. Given polynomial f(x) = x + x – 9x – 3x + 18 has two of its
Sol. Let p (x) = x + 2x + kx + 3.
We are given that when p (x) is divided by the linear zeros  3 and 3.
polynomial x – 3, the remainder is 21.
 p (3) = 21 (Remainder Theorem)  (x  3) (x  3) is a factor of f (x),
3 2
 3 + 2 × 3 + k × 3 + 3 = 21 2
i.e., x – 3 is a factor of f (x).
 27 + 18 + 3k + 3 = 21 Now, we apply the division algorithm to the given polynomial
 3k = 21 – 27 – 18 – 3 2
with x – 3.
 3k = –27
 k = –9
3 2 2
Hence, p (x) = x + 2x – 9x + 3. x +x–6
To find the quotient obtained on dividing p(x) by x–3, we x 2  3 x 4  x 3  9x 2  3x  18
perform the following division :
x4  3x 2
 
x 2  5x  6
x  3 x  2x 2  9x  3
3 x 3  6x 2  3x  18

x 3  3x 2 x3  3x
   

5x 2  9x  3 6x 2  18

5x 2  15x 6x 2  18
   
6x  3 0  Remainder
6x  18
  4 3 2
Thus, x + x – 9x – 3x + 18
21
2 2
= (x – 3) (x + x – 6)
2
Hence, p (x) = (x + 5x + 6) (x–3) + 21 2 2
= (x – 3) × {x + 3x – 2x – 6}
(Divisor × Quotient + Remainder) 2
3 2 2
= (x – 3) × {x (x + 3) – 2 (x + 3)}
 x + 2x – 9x + 3 – 21 = (x + 5x + 6) (x – 3) 2
3 2 2 = (x – 3) × (x + 3) (x – 2)
 x + 2x – 9x – 18 = (x + 3x + 2x + 6) (x –3)
3 2 Putting x + 3 = 0 and x – 2 = 0
 x + 2x – 9x – 18 = (x + 3) (x + 2) (x – 3)
Hence, the zeros of x + 2x – 5x – 18 are given by
3 2 we get x = –3 and x = 2, i.e., –3 and 2 are the other two zeros
x + 3 = 0, x + 2 = 0, x – 3 = 0 of the given polynomial.
 x = –3, – 2, 3 Hence  3, 3, –3, 2 are the four zeros of the given
3 2
 The zeros of x + 2x – 9x – 18 are – 3, –2, 3.
polynomial.
 QUADRATIC EQUATION 7

Example – 3 Example – 4

If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by

3x 2  7x  8
another polynomial x2 – 2x + k, the remainder comes out to Solve the inequality, 2
x2 1
be x + a, find k and a.

Sol. By division algorithm Sol. Domain : x  R

4 3 2 2 Given inequality is equivalent to
x – 6x + 16x – 25x + 10 = (x – 2x + k) q (x) + (x + a)

where q (x) is the quotient. 3x 2  7x  8

2 0
x2 1
As the degree on L.H.S. is 4; therefore, q (x) must be of
degree 2. 3x 2  7x  8  2x 2  2
 0
2 x2 1
Let q (x) = lx + mx + n, l 0.
4 3 2 2 2
Then x – 6x + 16x – 25x + 10 = (x – 2x + k) (lx + mx + n) + x + a 3x 2  7x  6  x  1 x  6 
 2
0  0
4 3 2 4 3 2
x 1 x2 1
 x – 6x + 16x – 25x + 10 = lx + (m–2l) x + (n–2m + kl) x + (mk
– 2n + 1) x + nk + a

Equating coefficients of like powers of x on the two sides,

we obtain
 x  [1, 6]
l=1 ... (1)
Example – 5
m – 2l = –6 ... (2)
Solve the equation 25x2 – 30x + 11 = 0 by using the general
n – 2m + kl = 16 ... (3) expression for the roots of a quadratic equation.
mk – 2 n + l = – 25 ... (4)
Sol. Comparing the given equation with the general form of a
and nk + a = 10 ... (5) 2
quadratic equation ax + bx + c = 0, we get
From (2), m = –6 + 2l = –6 + 2 × 1 = –4 and a = 25, b = –30 and c = 11.
Substituting these values in
then from (3), n = 16 + 2m – kl = 16 + 2 × (–4) – k × 1

 n=8–k ... (6)  b  b 2  4ac  b  b 2  4ac

 and  
2a 2a
From (4) and (6), we get
30  900  1100 30  900  1100
(–4) k – 2 (8 – k) + 1 = –25  and  
50 50
 – 4k – 16 + 2k = –25
30  200 30  200
 – 2k = –25 + 16 – 1   and  
50 50
 – 2k = –10 k = 5
30  10i 2 30  10i 2
Substituting this value of k in (6), we have   and  
50 50
n = 8 – 5 = 3 and then from (5),
3 2 3 2
   i and    i
we get 5 5 5 5
a = 10 – nk = 10 – 3 × 5 = –5. 3 2
Hence, the roots of the given equation are  i
5 5
 8 QUADRATIC EQUATION

Example – 6

Solve the following quadratic equation by factorization

method :
x2 – 5ix – 6 = 0 An equation of the form
f (x) f (x)
a +b =c
Sol. The given equation is where a, b, c  R
2
x – 5ix – 6 = 0 and a, b, c satisfies the condition a + b = c
2 2

2 2
 x – 5ix + 6i = 0 then solution of the equation is f (x) = 2 and no other

2
x – 3ix – 2ix + 6i = 0
2 solution of this equation.
2 2
(ii) Here, 3 + 5 = 34, then given equation has a solution
 x (x – 3i) – 2i (x – 3i) = 0
x–4=2
 (x – 3i) (x – 2i) = 0
 x = 6 is a root of the original equation
 x – 3i = 0, x – 2i = 0
 x = 3i, x = 2i
Hence, the roots of the given equation are 3i and 2i.
g(x)
An equation of the form {f (x)} is equivalent to the
Example – 7 equation
g(x) g (x) log f (x)
Solve the equation {f (x)} = 10 where f (x) > 0
x+1 2–x
(i) 15.2 + 15.2 = 135.
x–4 x–4
(iii) We have 5x x
8x 1  500
(ii) 3 +5 = 34
x 3 2
(iii) 5x x
8x 1  500  5x 8x 1 = 5 . 2

 x 1 
 
Sol. (i) The equation rewrite in the form  5x .8 x 
 53.22

3x  3
60
30.2 x   135  5x .2 x
 53.22
2x
 x 3 
x  
Let t = 2  5x 3.2 x 
1
2
then 30t – 135 t + 60 = 0  (5.2 )
1/x (x – 3)
=1
2
6t – 27t + 12 = 0 is equivalent to the equation
2
 6t – 24t – 3t + 12 = 0  x 3 log  5.21/ x 
10 1
 (t – 4) (6t – 3) = 0 1/x
 (x – 3) log (5.2 ) = 0
1 Thus original equation is equivalent to the collection of
then t1 = 4 and t2 =
2 equations
1/x
x – 3 = 0, log (5.2 ) = 0
thus given equation is equivalent to 1/x
 x = 3, 5.2 = 1
1/x
x x 1  2 = (1/5)
2 = 4 and 2 =
2  x = –log52
then x = 2 and x = – 1 Hence roots of the original equation are

Hence roots of the original equation are x1 = 2 and x2 = – 1 x = 3 and x = –log52

 QUADRATIC EQUATION 9

Example – 8 Example – 10

Form an equation whose roots are cubes of the roots of If P (x) = ax2 + bx + c and Q (x) = – ax2 + bx + c,
3 2
equation ax + bx + cx + d = 0
where ac  0, show that the equation

Sol. Replacing x by x1/3 in the given equation, we get P (x) · Q (x) = 0 has at least two real roots.

a (x1/3)3 + b (x1/3)2 + c (x1/3) + d =0 Sol. Roots of the equation P (x) Q (x) = 0

 ax + d = – (bx2/3 + cx1/3) ....... (i) i.e., (ax2 + bx + c) (–ax2 + bx + c) = 0 will be
 (ax + d)3 = – (bx2/3 + cx1/3)3 roots of the equations

 a3x3 + 3a2dx2 + 3ad2x + d3 ax2 + bx + c = 0 ....... (i)

= – [b3x2 + c3x + 3bcx (bx2/3 + cx1/3)] and – ax2 + bx + c = 0 ........ (ii)

 a3x3 + 3a2dx2 + 3ad2x + d3 If D1 and D2 be discriminants of (i) and (ii) then

[ – b3x2 – c3x + 3bcx (ax + d)] [From Eq. (i)] D1 = b2 – 4ac and D2 = b2 + 4ac

 a3x3 + x2 (3a2d – 3abc + b3) Now D1 + D2 = 2b2 0

+ x (3ad2 – 3bcd + c3) + d3 = 0 (since, b may be zero)

This is the requied equation. i.e., D 1 + D2  0

Example – 9 Hence, at least one of D1 and D2  0

i.e., at least one of the equations (i) and (ii) has real roots and
If , ,  be the roots of the equation
therefore, equation P (x) Q (x) = 0 has at least two real roots.
2 2
x (1 + x ) + x (6 + x) + 2 = 0,
Alternative Sol.
–1 –1 –1
then the value of  +  +  is
Since, ac  0

1  ac < 0 or ac > 0
(a) –3 (b)
2
Case I :
1
(c)  (d) None of these If ac < 0  – ac > 0
2
then D1 = b2 – 4ac > 0
Sol. 2x3 + 6x2 + x + 2 = 0 has roots , , .
Case II :
So, 2x3 + x2 + 6x + 2 = 0 has roots –1, –1, –1
(writing coefficients in revers order, since roots are reciprocal) If ac > 0

then D2 = b2 + 4ac > 0

Hence, Sum of the roots = –
 Coefficient of x  2

 Coefficient of x  3
So, at least one of D1 and D2 > 0.

1 Hence, at least one of the equations (i) and (ii) has real roots.
  1  1   1  
2
Hence, equation P (x) · Q (x) = 0 has at least two real roots.
Hence, (c) is the correct answer.
 10 QUADRATIC EQUATION

Example – 11 Example – 13

Find all roots of the equation x4 + 2x3 – 16x2 – 22x + 7 = 0 if one

If x2 – ax + b = 0 and x2 – px + q = 0 have a root in common
and the second equation has equal roots. root is 2  3 .
ap
show that b + q = .
2 Sol. All coefficients are real, irrational roots will occur in
conjugate pairs.
Sol. Given equations are x2 – ax + b = 0 ...(i)
Hence another roots is 2  3 .
and x2 – px + q = 0 ...(ii)
 Product of these roots = (x – 2 – 3 ) (x – 2 + 3 )
Let  be the common root. Then roots of Eq. (ii) will be  2
and . Let  be the other root of Eq. (i). Thus roots of Eq. = (x – 2) – 3
2
(i) are ,  and those of Eq. (ii). are ,  = x – 4x + 1
= a ...(iii) 4 3 2 2
b ...(iv) Dividing x + 2x – 16x – 22x + 7 by x – 4x + 1 then the other
2
2p ...(v) quadratic factor is x + 6 x + 7
 q ...(vi) then the given equation reduce in the form
LHS = b + q = + 2 =  (+ ) ...(vii) 2 2
(x – 4x + 1) (x + 6x + 7) = 0
ap      2
and RHS =       ...(viii) 
2
x + 6x + 7 = 0
2 2
From Eqs. (vii) and (viii), LHS = RHS
6  36  28
Example – 12 then x 
2

The diagram shows the graph of = 3  2

y = ax2 + bx + c, Then,
Hence roots 2  3,  3  2
(0, c)
Example – 14

x1 x2 x 2  3x  4
If x is real, then prove that the values of lies
x 2  3x  4
(a) a > 0 (b) b < 0 1
(c) c > 0 (d) b2 – 4ac = 0 between and 7.
7

Sol. As it is clear from the figure that it is a parabola opens

downwards i.e. a < 0.
x 2  3x  4
 It is y = ax2 + bx + c i.e. degree two polynomial Sol. Let y = . Then,
x 2  3x  4
Now, if ax2 + bx + c = 0
 it has two roots x1 and x2 as it cuts the axis at two distinct x2 (y – 1) + 3x (y + 1) + 4 (y – 1) = 0
point x1 and x2.  for x  real, D  0
Now from the figure it is also clear that x1 + x2 < 0  9 (y + 1)2 – 16 (y – 1)2 0
(i.e. sum of roots are negative)
 – 7y2 + 50y – 7  0
b b  (7y – 1) (y – 7)  0
 0  0
a a
 b < 0 ( a < 0) (b) is correct. 1
Hence, the given expression lies between and 7.
As the graph of y = f (x) cuts the + y-axis at (0, c) 7

where c > 0  (c) is correct.

 QUADRATIC
11 EQUATION 11
QUADRATIC EQUATION

Example – 15 As in Case I, [1, 3] should be the subset of

a, b, c  R, a  0 and the quadratic equation (a + 3, 3a)

ax2 + bx + c = 0 has no real roots, then, i.e., a+3<1 and 3a > 3 ...(iv)
Combining (iii) and (iv), we get :
(a) a + b + c > 0 (b) a (a + b + c) > 0
(c) b (a + b + c) > 0 (d) c (a + b + c) >0 a   i.e. No solution ...(v)

Combining both cases, we get : a  (0, 1/3)

Sol. Let f (x) = ax2 + bx + c. It is given that f (x) = 0 has no real
Alternate Sol.
roots. So, either f (x) > 0 for all x  R
Let f (x) = (x – 3a) (x – a – 3)
or f (x) < 0 for all x  R i.e. f (x) has same sign for all values
of x. for given equality to be true for all values of
x  [1, 3], 1 and 3 should lie between the roots of
 f (0) f (1) > 0
f (x) = 0.
 c (a + b + c) > 0
 f (1) < 0 and f (3) < 0
Also, af (1) > 0
Consider f (1) < 0 :
 a (a + b + c) > 0.
 (1 – 3a) (1 – a – 3) < 0
Example – 16  (3a – 1) (a + 2) < 0
Find the values of a for which the inequality  a (–2, 1/3) ...(i)
(x – 3a) (x – a – 3) < 0 is satisfied for all x such that Consider f (3) < 0 :
1  x  3.
 (3 – 3a) (3 – a – 3) < 0
Sol. (x – 3a) (x – a – 3) < 0  (a – 1) (a) < 0
 a  (0, 1) ...(ii)
Case I :
Combining (i) and (ii), we get :
Let 3a < a + 3 a < 3/2 ...(i)
a  (0, 1/3)
Solution set of given inequality is x (3a, a + 3)
Now for given inequality to be true for all Example – 17
x  [1, 3], set [1, 3] should be the subset of
(3a, a + 3) i.e. 1 and 3 lie inside 3a and a + 3 on number line. If ax2 – bx + 5 = 0 does not have two distinct real roots,
then find the minimum value of 5a + b.
So we can take, 3a < 1 and a + 3 > 3 ...(ii)
Combining (i) and (ii), we get : Sol. Let f(x) = ax2 – bx + 5
 a (0, 1/3) Since, f (x) = 0 does not have two distinct real roots, we
Case II : have either
Let 3a > a + 3  a > 3/2 ...(iii) f  x   0  x R or f  x   0  x R
Solution set of given inequality is
But f (0) = 5 > 0, so f (x)  0  x R
x (a + 3, 3a)
In particular f (–5) > 0  5a + b > – 1
Hence, the least value of 5a + b is –1.
 QUADRATIC
12 EQUATION 12
QUADRATIC EQUATION

EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS

Basics, Sum & Product of Roots
10. If 9x 2  6x  1   2  x  , then
1. Roots of the equation x2 + x (2 – p2) – 2p2 = 0 are
(a) – p2 and – 2 (b) p2 and – 2  3 1  3 1
(a) x    ,  (b) x    , 
(c) – p2 and 2 (d) p2 and 2  2 4  2 4

x  1  x  1  1, then x is equal to  3 1 1
2. If (c) x    ,  (d) x 
 2 4 4
(a) 2/3 (b) 3/5
(c) 5/4 (d) 4/5 11. If  and  are the roots of ax 2 + bx + c = 0 then the value

of (aα + b)-2 + (aβ + b)-2 is equal to

x y 10
3. If   and x + y = 10, then the value of xy will
y x 3
b 2 - 2ac c2 - 2ab
(a) (b)
be a 2 c2 a 2 b2
(a) 36 (b) 24
(c) 16 (d) 9 a 2 - 2bc
(c) (d) None
b2c2
8x 2 16x  51
4.  3 if x is such that
 2x  3 x  4  12. If  and  are the roots of 4x 2 + 3x + 7 = 0, then the

(a) x < – 4 (b) –3 < x < 3/2 1 1

value of 3
+ is
(c) x > 5/2 (d) all these true α β3

5. The sum of all real roots of the equation

27 63
|x – 2|2 + |x – 2| – 2 = 0, is (a)  (b)
64 16
(a) 0 (b) 8
(c) 4 (d) none of these 225
(c) (d) None of these
6. 2
The equation x –3 |x| + 2 = 0 has 343

(a) no real roots (b) one real root 13. If  and  are the roots of the equation

(c) two real roots (d) four real roots x 2  px  p2  q  0, then the value of

7. The sum of the real roots of the equation x2 + |x| – 6 = 0 is  2    2  q 

(a) 4 (b) 0
(a) 0 (b) 1
(c) – 1 (d) none of these
(c) q (d) 2q
8. The product of the real roots of the equation
|2x + 3|2 – 3 |2x + 3| + 2 = 0, is 14. If  ,  are the roots of the equation 8x 2  3x  27  0,

(a) 5/4 (b) 5/2 1/3 1/3

 2   2 
then the value of   
  is
(c) 5 (d) 2     

9. The roots of the equation |x2 – x – 6| = x + 2 are (a) 1/4 (b) 1/3

(a) –2, 1, 4 (b) 0, 2, 4 (c) 7/2 (d) 4

(c) 0, 1, 4 (d) –2, 2, 4

QUADRATIC EQUATION 13

24. If , are the roots of the equation x2 – p (x + 1) –c =0, then

15. If p, q are the roots of the equation x 2  px  q  0 where
(+ 1) (+ 1) =
both p and q are non-zero, then (p, q) =
(a) c (b) c–1
(a) (1, 2) (b) (1, –2)
(c) 1–c (d) none of these
(c) (–1, 2) (d) (–1, –2)
16. The product of the roots of the equation 25. If  and  are the roots of x 2  p (x  1)  c  0, then the

mx 2  6x  (2m  1)  0 is –1. Then m is equal to  2  2  1 2  2  1

value of  is
 2  2  c 2  2  c
(a) 1 (b) 1/3
(c) –1 (d) –1/3 (a) 2 (b) 1
17. If one root of the equation ax2 + bx + c = 0, a  0, is reciprocal (c) –1 (d) 0
of the other, then
26. If one root of x 2  x  k  0 be square of the other, then k
(a) b = c (b) a = c
is equal to
(c) a = 0 (d) b = 0
(a) 2  3 (b) 3  2
18. If the sum of the roots of the equation ax2 + 2x + 3a = 0 is
equal to their product, then value of a is (c) 2  5 (d) 5  2
2
(a)  (b) –3 27. The number of values of a for which
3
1 (a2 – 3a + 2) x2 + (a2 – 5a + 6) x + a2 – 4 = 0
(c) 4 (d) 
2 is an identity in x, is
19. If the product of the roots of the quadratic equation
mx2 – 2x + (2m – 1) = 0 is 3 then the value of m (a) 0 (b) 2

(a) 1 (b) 2 (c) 1 (d) 3

(c) –1 (d) 3 28. If p (x  1)2  q (x 2  3x  2)  x  1  0 be an identity in

20. If the roots of px2 + qx + 2 = 0 are reciprocals of each other, x, then p, q are
then (a) 2, –2 (b) 1, –1
(a) p = 0 (b) p = –2 (c) 0, 0 (d) none
(c) p = ± 2 (d) p = 2 29. If the difference between the roots of x2 + ax + b = 0 and
2
21. If the equation (k – 2) x – (k – 4) x – 2 = 0 has difference of x2 + bx + a = 0 is same and a  b, then
roots as 3 then the value of k is (a) a + b + 4 = 0 (b) a + b –4 = 0
(a) 1, 3 (b) 3, 3/2 (c) a – b –4 = 0 (d) a – b + 4 = 0
(c) 2, 3/2 (d) 3/2, 1 30. 2
If roots of the equation x + ax + 25 = 0 are in the ratio of
22. If sin  and cos are the roots of the equation 2 : 3 then the value of a is

lx 2  mx  n  0, then 5  25
(a) (b)
6 6
(a) l 2 – m 2  2ln  0 (b) l 2  m 2  2ln  0
5
(c) l 2 – m 2 – 2ln  0 (d) l 2  m 2 – 2ln  0 (c) (d) none of these
6
23. The roots of the equation x 2  px  q  0 are 31. If the roots of the equations x2 + 3x + 2 = 0 & x2 – x +  = 0
tan 22º and tan 23º then are in the same ratio then the value of  is given by
(a) p + q = 1 (b) p + q = –1 (a) 2/7 (b) 2/9
(c) p – q = 1 (d) p – q = –1 (c) 9/2 (d) 7/2
 14 QUADRATIC EQUATION

32. If  are roots of the equation x2 – 5x + 6 = 0 then the 40. If , ,  are the roots of the equation
equation whose roots are  + 3 and  + 3 is
(a) x2 – 11x + 30 = 0 (b) (x – 3)2 – 5 (x – 3) + 6 = 0 x3  px 2  qx  r  0, then
(c) Both (a) and (b) (d) None of these
(1  2 ) (1  2 ) (1   2 ) is equal to
33. If  are roots of Ax + Bx + C = 0 and  are roots of
2

x2 + px + q = 0, then p is equal to (a) (1  q) 2  (p  r) 2 (b) (1  q) 2  (p  r)2

2 2 2 2
(a) (B – 2AC)/A (b) (2AC – B )/A
2
(c) (B – 4AC)/A 2
(d) (4AC – B2)A2 (c) (1  q) 2  (p  r)2 (d) none of these

34. If ,  are roots of the equation 41. If (x2 – 3x + 2) is a factor of x4 – px2 + q = 0, then the values
of p and q are
 2 2 (a) –5, 4 (b) 5, 4
ax 2  3x  2  0 (a  0), then  is greater than
  (c) 5, –4 (d) –5, –4
(a) 0 (b) 1 42. The least integral value of k which makes the roots of the
equation x2 + 5x + k = 0 imaginary is
(c) 2 (d) none of these
(a) 4 (b) 5
35. In a quadratic equation with leading coefficient 1, a student
reads the coefficient 16 of x wrongly as 19 and obtain the (c) 6 (d) 7
roots as –15 and –4. The correct roots are 43. The roots of the quadratic equation 7x2 – 9x + 2 = 0 are
(a) 6, 10 (b) –6, –10 (a) Rational and different (b) Rational and equal
(c) –7, –9 (d) none of these (c) Irrational and different (d) Imaginary and different
Cubic, Biquadratic, Nature of Roots
44. The roots of the equation x 2  2 2 x 1  0 are
36. If   and  are the roots of the cubic equation
(x – 1)(x2 + x + 3)=0, then the value of  +  +  is equal to (a) Real and different (b) Imaginary and different
(a) –1 (b) 0 (c) Real and equal (d) Rational and different
(c) 2 (d) 3 45. If a, b, c,  Q and b + c  0 then the roots of the equation

37. If one root of equation x2 + ax + 12 = 0 is 4 while the equation (b + c) x2 – (a + b + c) x + a = 0 are

x2 + ax + b = 0 has equal roots, then the value of b is (a) Real (b) Real and Rational
4 49 (c) Non real and different (d) Real and equal
(a) (b)
49 4 46. If l, m, n are real, l  m, then the roots of the equation
(l – m) x2 – 5 (l + m) x – 2 (l – m) = 0 are
7 4
(c) (d) (a) real and equal (b) Non real
4 7
(c) real and unequal (d) none of these
38. If , ,  are the roots of the equation 2x3 – 3x2 + 6x + 1 = 0, then
47. If a,b,c are distinct real numbers then the equation
2 + 2 + 2 is equal to
(b  c) x 2  (c  a) x  (a  b)  0 has
15 15
(a)  (b)
4 4 (a) equal roots (b) irrational roots

9 (c) rational roots (d) none of these

(c) (d) 4
4 48. If a,b,c are distinct rational numbers then roots of equation
39. The value of m for which the equation (b  c  2a) x 2  (c  a  2b)x  (a  b  2c)  0 are

x 3  mx 2  3x  2  0 has two roots equal in (a) rational (b) irrational

magnitude but opposite in sign, is (c) non-real (d) equal
(a) 1/2 (b) 2/3
(c) 3/4 (d) 4/5
QUADRATIC
15 EQUATION 15
QUADRATIC EQUATION

49. If a,b,c are distinct rational numbers and a + b + c = 0, then Common Root
the roots of the equation
57. If the equations x 2  2x  3  0 and
2 2 2 2
(c  ab) x  2 (a  bc) x  (b  ac)  0 are
2x 2  3x  5  0 have a non-zero common root, then
(a) imaginary (b) real and equal

(c) real and unequal (d) none of these
(a) 1 (b) –1
50. If a,b,c are distinct rational numbers and a + b + c = 0, then
(c) 3 (d) None
the roots of the equation
58. The value of a so that the equations
(b  c  a) x 2  (c  a  b) x  (a  b  c)  0 are
(2a  5) x 2  4x  15  0 and
(a) imaginary (b) real and equal
(c) real and unequal (d) none of these (3a  8) x 2  5x  21  0 have a common root, is
51. If a  Z and the equation (x – a) (x – 10) + 1 = 0 has integral (a) 4, 8 (b) 3, 6
roots, then the values of a are (c) 1, 2 (d) None
(a) 8, 10 (b) 10, 12
59. If a,b,c  R, the equation ax2  bx  c  0 (a,c  0) and
(c) 12, 8 (d) none
52. The quadratic equation with rational coefficients whose x 2  2x  3  0 have a common root, then a : b : c =
one root is 2 + 3 is (a) 1 : 2 : 3 (b) 1 : 3 : 4
(a) x2 – 4x + 1 = 0 (b) x2 + 4x + 1 = 0 (c) 2 : 4 : 5 (d) None
(c) x2 + 4x – 1 = 0 (d) x2 + 2x + 1 = 0
60. If the equations k(6x 2  3)  rx  2x 2  1  0 and
53. The quadratic equation with real coefficients whose one
root is 2 – i 3 is 6k (2x 2  1)  px  4x 2  2  0 have both roots common,
(a) x2 – 4x + 7 = 0 (b) x2 + 4x – 7 = 0 then the value of (2r – p) is

(c) x2 – 4x – 7 = 0 (d) none of these (a) 0 (b) 1/2

54. The equation of the smallest degree with real coefficients (c) 1 (d) None of these
having 1 + i as one of the roots is Range of Rational Expression
(a) x2 + x + 1 = 0 (b) x2 – 2x + 2 = 0
x 2  2x  4
61. If x is real, then takes values in the interval
x 2  2x  4
(c) x2 + 2x + 2 = 0 (d) x2 + 2x –2 = 0

55. If  are the roots of a quadratic equation x2 – 3x + 5 =0 then 1  1 

(a)  ,3 (b)  ,3 
the equation whose roots are ( 2 – 3 + 7) and 3  3 
(2 – 3 + 7) is
(a) x2 + 4x + 1 = 0 (b) x2 – 4x + 4 = 0  1 
(c) (3, 3) (d)   ,3 
 3 
(c) x2 – 4x – 1 = 0 (d) x2 + 2x + 3 = 0
56. If the roots of a1 x2 + b1 x + c1 = 0 are 1, 1 and those of x 2  14x  9
a2x2 + b2 x + c2 = 0 are 2 2 such that 1 2 = 1 2 = 1, then 62. If x is real then the value of the expression
x 2  2x  3
a1 b1 c1 a1 b1 c1 lies between
(a)   (b)  
a 2 b2 c2 c2 b 2 a 2 (a) –3 and 3 (b) –4 and 5
(c) a1a2 = b1b2 = c1c2 (d) none of these (c) –4 and 4 (d) –5 and 4
 16 QUADRATIC EQUATION

Graph of Quadratic Expression 66. If a, b, c  R, for which of the following graphs of

63. 2
If a, b  R & the quadratic equation ax – bx + 1 = 0 has y = ax2 + bx + c, the product a b c is negative.
imaginary roots then a + b + 1 is
(a) positive
(b) negative
(c) zero
(d) depends on the sign of b (a) (b)
64. If a, b, c  R, the graph of the quadratic polynomial;
y = ax2 + bx + c is as shown in the figure. Then

(c) (d)
x

(a) b2 – 4ac > 0 (b) b < 0 67. The integer k for which the inequality
(c) a > 0 (d) c < 0 x2 –2 (4k–1) x + 15k2 – 2k – 7 > 0 is valid for any x, is
65. If a, b, c  R, Which of the following graph represents, (a) 2 (b) 3
f (x) = ax2 + bx + c when a > 0, b < 0 and c < 0 ?
(c) 4 (d) none of these

68. If a  R and x2 + 2ax + 10 –3a > 0 for all x R, then

(a) –5 < a < 2 (b) a < –5
(a) (b)
(c) a > 5 (d) 2 < a < 5
69. The real values of ‘a’ for which (a2 –1) x2 + 2 (a –1) x + 2 is
positive for any x, are
(a) a 1 (b) a 1
(c) a > –3 (d) a < –3 or a  1
70. The real value of a for which the sum of the squares of the
roots of the equation x2 – (a – 2) x – a – 1 = 0 assumes the
(c) (d) least value, is
(a) 0 (b) 1
(c) 2 (d) 3
71. If a, b, c R. Both roots of the equation
(x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are
(a) positive (b) negative
(c) real (d) imaginary
 QUADRATIC EQUATION 17

72. If a < b < c < d, then roots of 76. The value of ‘a’ for which the equation
(x – a) (x – c) + 2 (x – b) (x – d) = 0 are
x 2  2 (a  1) x  (2a  1)  0 has both roots
(a) real and equal (b) real and unequal
positive is
(c) imaginary (d) rational
(a) a > 0 (b) 0 < a < 4
Location of Roots
(c) a  4 (d) None of these
73. The value of k for which the equation
3x2 + 2x (k2 + 1) + k2 – 3k + 2 = 0 77. If the equation x 2  2 (a  1) x  9a  5  0 has only
has roots of opposite signs, lies in the interval negative roots, then
(a) (–, 0) (b) (–, –1) 5 
(a) a  (, 6) (b) a   , 1  (6, )
(c)(1, 2) (d) (3/2, 2) 9 
74. The value of a for which the equation
(c) a  (0, 6) (d) a  0
2
2x  2 (2a  1)x  a (a  1)  0 has roots,  and  such 78. The value of k for which both the roots of the
that   a   is equation 4x 2  20kx  (25k 2  15k  66)  0 are less
(a) a  0 (b) a < 0 then 2, lies in

(c) 3  a  0 (d) None of these (a) (4/ 5, 2) (b) (2, 0)

75. The value of for which (c) (–1, –4/5) (d) (, 1)

2x 2  2 (2  1) x   (  1)  0 may have one root less 79. If the roots of x 2  x  a  0 exceed a, then
than and other root greater than are given by (a) 2 < a < 3 (b) a > 3
(a) 1    0 (b) –1    0 (c) –3 < a < 3 (d) a < – 2

(c)   0 (d)   0 or   1 80. The range of values of m for which the equation

(m  5) x 2  2 (m  10) x  m  10  0 has real roots of

the same sign, is given by
(a) m > 10 (b) –5 < m < 5
(c) m < –10, 5 < m  6 (d) None of these
 QUADRATIC
18 EQUATION 18
QUADRATIC EQUATION

EXERCISE - 2 : PREVIOUS YEAR JEE MAINS QUESTIONS

1. If    but a2 = 5 – 3 and 2 = 5 – 3 then the equation 9. If (1 – p) is a root of quadratic equation x2 + px + (1 – p) = 0, then

its roots are (2004)
whose roots are / and / is (2002)
2 2 (a) 0, –1 (b) –1, 1
(a) 3x – 25x + 3 = 0 (b) x + 5x – 3 = 0
(c) 0, 1 (d) –1, 2
(c) x2 – 5x + 3 = 0 (d) 3x2 – 19x + 3 = 0
2
10. If one root of the equation x + px + 12 = 0 is 4, while the
2. Difference between the corresponding roots of
equation x2 + px + q = 0 has equal roots,
x2 + ax + b = 0 and x2 + bx + a = 0 is same and a  b , then
then the value of q is (2004)
(2002)
(a) 3 (b) 12
(a) a + b + 4 = 0 (b) a + b – 4 = 0
(c) 49/4 (d) 4
(c) a – b – 4 = 0 (d) a – b + 4 = 0
11. The value of a for which the sum of the squares of the
3. Product of real roots of the equation
roots of the equation x2 – (a – 2)x – a – 1 = 0 assume the
x2 + |x| + 9 = 0 (2002) least value, is (2005)
(a) is always positive (b) is always negative (a) 0 (b) 1
(c) does not exist (d) none of the above (c) 2 (d) 3
2
4. If p and q are the roots of the equation x + px + q = 0, then 12. If the roots of the equation x2 – bx + c = 0 be two
(2002) consecutive integers, then b2 – 4c equals (2005)
(a) p = 1, q = – 2 (b) p = 0, q = 1 (a) 3 (b) –2
(c) p = –2, q = 0 (d) p = –2, q = 1 (c) 1 (d) 2
5. If a, b, c are distinct real numbers and a2 + b2 + c2 = 1, 13. If both the roots of the quadratic equation
then ab + bc + ca is (2002) x2 – 2kx + k2 + k – 5 = 0 are less than 5,
(a) less than 1 (b) equal to 1 then k lies in the interval (2005)
(c) greater than 1 (d) any real no
(a)  6,  (b) (5, 6]
6. The value of a for which one root of the quadratic equation
(a2 – 5a + 3) x2 + (3a – 1) x + 2 = 0 is twice as large as the (c) [4, 5] (d)  ,4 
other, is (2003)
14. If the roots of the quadratic equations x2 + px + q = 0 are
(a) –2/3 (b) 1/3
tan 30o and tan 15o respectively, then the value of 2 + q – p is
1 2 (2006)
(c)  (d)
3 3 (a) 2 (b) 3
7. The number of real solution of the equations x2 – 3|x| + 2 = 0 is (c) 0 (d) 1
(2003) 15. All the values of m for which both roots of the equation
(a) 4 (b) 1 x2 – 2mx + m2 –1 = 0 are greater than –2 but less than 4,
lie in the interval (2006)
(c) 3 (d) 2
(a) –2 < m < 0 (b) m > 3
8. Let two numbers have arithmetic mean 9 and geometric mean
4. Then these numbers are the roots of the quadratic equation (c) –1 < m < 3 (d) 1 < m < 4
(2004) 16. If the difference between the roots of the equation
(a) x2 + 18x – 16 = 0 (b) x2 – 18x + 16 = 0 x2 + ax + 1 = 0 is less than 5 , then the set of possible
(c) x2 + 18x + 16 = 0 (d) x2 – 18x – 16 = 0. values of a is (2007)
(a)  3,  (b)  , 3

(c)  3, 3 (d)  3,  

QUADRATIC
19 EQUATION 19
QUADRATIC EQUATION

17. The quadratic equations 23. If a R and the equation –3 (x – [x])2 + 2 (x – [x]) + a2 = 0
x2 – 6x + a = 0 (where [x] denotes the greatest integer x) has no integral
and 2
x – cx + 6 = 0 solution, then all possible values of a lie in the interval:
(2014)
have one root in common. The other roots of the first and
second equations are integers in the ratio 4 : 3. Then the (a) (–, –2) (2,  (b) (–1, 0) (0, 1)
common root is (2008) (c) (1, 2) (d) (–2, –1)
(a) 2 (b) 1 24. If equations ax2 + bx + c = 0, (a, b  R, a  0) and
(c) 4 (d) 3 2x2 + 3x + 4 = 0 have a common root then a : b : c equals:

18. If the roots of the equation bx2 + cx + a = 0 be imaginary, (2014/Online Set–1)

then for all real values of x, the expression (a) 1 : 2 : 3 (b) 2 : 3 : 4
3b2x2 + 6bcx +2c2 is (2009)
(c) 4 : 3 : 2 (d) 3 : 2 : 1
(a) greater than 4ab (b) less than 4ab
(c) greater than –4ab (d) less than –4ab 1 1
25. If and are the roots of the equation,
 
19. If  and are the roots of the equation x2 – x +1 = 0, then
2009 + 2009 is equal to (2010) ax2 + bx + 1 = 0
(a) –2 (b) –1 (a  0, a, b  R), then the equation,
(c) 1 (d) 2 x (x + b3) + (a3 - 3abx) = 0 has roots:

20. Sachin and Rahul attempted to solve a quadratic equation. (2014/Online Set–1)
Sachin made a mistake in writing down the constant term
and ended up in roots (4, 3). Rahul made a mistake in (a) 3/2 and 3/2 (b) 1/2 and 1/2
writing down coefficient of x to get roots (3, 2). The correct
roots of equation are (2011) (c)  and  (d) 3/2 and 3/2

(a) –4, –3 (b) 6, 1 26. If  and  are roots of the equation,

(c) 4, 3 (d) –6, –1
x2 -4 2 kx + 2e4 ln k - 1 = 0 for some k, and  2  2  66
sin x – sin x
21. The equation e –e – 4 = 0 has (2012)
(a) infinite number of real roots then 3  3 is equal to: (2014/Online Set–2)

(b) no real roots (a) 248 2 (b) 280 2

(c) exactly one real root
(c) 32 2 (d) 280 2
(d) exactly four real roots
22. Let  and  be the roots of equation 27. The sum of the roots of the equation, x2 + |2x - 3| - 4 = 0, is
(2014/Online Set–3)
1 1
px2 + qx + r = 0, p 0. If p, q, r are in A.P. and   4, (a) 2 (b) -2
 
then the value of |–| is : (2014) (c) 2 (d)  2

2 13 61 28. The equation 3x 2  x  5  x  3, where x is real, has:

(a) (b)
9 9 (2014/Online Set–4)
(a) no solution
2 17 34
(c) (d) (b) exactly one solution
9 9
(c) exactly two solutions
(d) exactly four solutions
 20 QUADRATIC EQUATION

29. Let  and  be the roots of equation x2 – 6 x –2 = 0. 36. Let p(x) be a quadratic polynomial such that p(0) = 1. If
a10  2a 8 p(x) leaves remainder 4 when divided by x “ 1 and it leaves
If an = n –n, for n1, then the value of is remainder 6 when divided by x + 1; then :
2a 9
(2017/Online Set–1)
equal to: (2015)
(a) p(2) = 11 (b) p(2) = 19
(a) 3 (b) –3
(c) p(–2) = 19 (d) p (–2) = 11
(c) 6 (d) –6
37. The number of real values of  for which the system of
30. If 2 + 3i is one of the roots of the equation linear equations
2x3 – 9x2 + kx – 13 = 0, k  R, then the real root of this
equation (2015/Online Set–1) 2x + 4y – z = 0
(a) exists and is equal to 1 4x + y + 2z = 0
x + 2y + 2z = 0
1
(b) exists and is equal to – has infinitely many solutions, is (2017/Online Set–1)
2
(a) 0 (b) 1
1 (c) 2 (d) 3
(c) exists and is equal to
2 38. The sum of all the real values of x satisfying the equation
2  5x  50)
(d) does not exist 2(x 1) ( x  1 is : (2017/Online Set–2)
31. A value of b for which the equations (a) 16 (b) 14
x2 + bx - 1 = 0, x2 + x + b = 0, have one root in common is (c) –4 (d) –5
(2015/Online Set–2)
39. If   R is such that the sum of the cubes of the roots of
(a)  2 (b) i 3 the equation, x2 +  2 -   x +  10 -   = 0 is minimum, then
(c) i 5 (d) 2 the magnitude of the difference of the roots of this
equation is : (2018/Online Set–1)
32. The sum of all real values of x satisfying the equation
x 2  4 x  60 (a) 4 2 (b) 2 5
x 2
 5x  5  = 1 is : (2016)

(a) –4 (b) 6 (c) 2 7 (d) 20

(c) 5 (d) 3 40. If f(x) is a quadratic expression such that

f(1) + f(2) = 0, and -1 is a root of f(x) = 0, then the other root
33. If b  C and the equations x 2 + bx –1 = 0 and of f(x) = 0 is : (2018/Online Set–2)
x2 + x + b = 0 have a common root different from –1, then
|b| is equal to : (2016/Online Set–1) 5 8
(a) - (b) -
(a) 2 (b) 2 8 5
(c) 3 (d) 3 5 8
(c) (d)
34. If x is a solution of the equation, 2x  1 8 5
 1 41. Let p, q and r be real numbers  p  q,r  0  , such that the
 2x  1  1,  x   , then
 2 4x 2  1 is equal to :
(2016/Online Set–2) 1 1 1
roots of the equation x + p + x + q = r are equal in
3 1
(a) (b) magnitude but opposite in sign, then the sum of squares
4 2
of these roots is equal to : (2018/Online Set–3)
(c) 2 (d) 2 2
35. If, for a positive integer n, the quadratic equation, p2 + q2
(a) (b) p2+ q2
2
x(x + 1) + (x + 1) (x + 2)+….+(x+ n  1 ) (x+n) = 10n
(c) 2(p2+ q2) (d) p2+q2+r2
has two consecutive integral solutions, then n is equal
to: (2017)
(a) 12 (b) 9
(c) 10 (d) 11
 QUADRATIC EQUATION 21

EXERCISE - 3 : ADVANCED OBJECTIVE QUESTIONS

1. The number of real values of the triplet (a, b, c) for which
8x 2 16x  51
a cos 2x + b sin2 x + c = 0 is satisfied by all real x, is 7.  3 if x is such that
 2x  3 x  4 
(a) 0 (b) 2
(c) 3 (d) infinite (a) x < – 4 (b) –3 < x < 3/2

2. cos  is a root of the equation 25x2 + 5x – 12 = 0, –1 < x < 0, (c) x > 5/2 (d) all these true

then the value of sin 2 is 8. Let  be the roots of ax2 + bx + c = 0 and ,  be the roots
of px2 + qx + r = 0 ; and D1, D2 the respective Discriminants
(a) 24/25 (b) –12/25
of these equations. If  are in A.P., then D1 : D2
(c) –24/25 (d) 20/25
3. Set of all values of x satisfying the inequality a2 a2
(a) (b)
p2 b2
x 2 – 7x  6 > x + 2 is

b2 c2
(c) (d)
 2 2  q2 r2
(a) x   – ,  (b) x   ,  
 11   11 
9. If 0  x  , then the solution of the equation
(c) x  (–, 1] [6, ) (d) x  [6, ) 2x 2x
16sin + 16cos = 10 is given by x equal to
4. If the product of the roots of the quadratic equation
mx2 – 2x + (2m – 1) = 0 is 3 then the value of m    
(a) , (b) ,
(a) 1 (b) 2 6 3 3 2

(c) –1 (d) 3
 
(c) , (d) none of these
6 2
 x  1 A B
5. If   , then 16A + 9B is
 2x  1 3x  1  2x  1  3x  1 10. Two real numbers  and  are such that +  = 3 and
equal to | – | = 4, then  and  are the roots of the quadratic
(a) 4 (b) 5 equation

(c) 6 (d) 8 (a) 4x2 – 12x – 7 = 0 (b) 4x2 – 12x + 7 = 0

(c) 4x2 – 12x + 25 = 0 (d) none of these
1 1 1
6. If a, b, c R. For the equation   , 11. 2
If (x + 1) is greater then 5x – 1 and less than 7x – 3 then the
xa xb xc
integral value of x is equal to
if the product of roots is zero, then the sum of roots if
(a) 1 (b) 2
b  c  0 is
(c) 3 (d) 4

2ab 12. The value of m for which one of the roots of

(a) 0 (b) x2 – 3x + 2m = 0 is double of one of the roots of
bc
x2 – x + m = 0 is
2bc 2bc (a) 0, 2 (b) 0, –2
(c) (d)
bc bc (c) 2, –2 (d) none of these
 22 QUADRATIC EQUATION

13. If and are the roots of the equation, x3 – x – 1 = 0 21. If  are the roots of the equation ax2 + bx + c = 0 and
1  1  1  Sn = n +n, then a Sn + 1 + c Sn – 1 =
then,   has the value equal to
1  1  1  (a) b Sn (b) b2Sn
(c) 2bSn (d) – bSn
(a) zero (b) –1
(c) –7 (d) 1 22. If the roots of equation x2 + bx + ac = 0 are  and roots
of the equation x2 + ax + bc = 0 are  then the value of
14. If the quadratic equations, 3x 2 + ax + 1 = 0 and
 respectively
2x2 + bx + 1 = 0 have a common root, then the value of the
expression 5ab – 2a2 – 3b2 is (a) a, b, c (b) b, c, a

(a) 0 (b) 1 (c) c, a, b (d) none of these

(c) –1 (d) none 23. If the roots of the equation, x3 + Px2 + Qx – 19 = 0 are each
one more than the roots of the equation,
15. If a, b, p, q are non-zero real numbers, the two equations,
2 a2x2 – 2 abx + b2 = 0 and p2x2 + 2 pqx + q2 = 0 have x3–Ax2+Bx–C=0 where A, B, C, P and Q are constants then
the value of A + B + C =
(a) no common root
(a) 18 (b) 19
(b) one common root if 2 a2 + b2 = p2 + q2
(c) 20 (d) none
(c) two common roots if 3 pq = 2 ab
24. If  are the roots of the equation ax2 + bx + c = 0, then
(d) two common roots if 3 qb = 2 ap
the equation ax2 – bx (x – 1) + c (x – 1)2 = 0 has roots
6x 2  5x  3
16. If  4, the least and the highest values of  
x 2  2x  6 (a) , (b)  – 1,  – 1
1–  1– 
4 x2 are
(a) 0 and 81 (b) 9 and 81   1–  1– 
(c) , (d) ,
(c) 36 and 81 (d) none of these  1 1  

17. If p & q are roots of the equation x2 – 2x + A = 0 and r & s 25. x2 + x + 1 is a factor of ax3 + bx2 + cx + d = 0, then the real
be roots of the equation x2 – 18 x + B = 0 if p < q < r < s be
root of above equation is (a, b, c, d  R)
in A.P., then A and B are respectively
(a) –d/a (b) d/a
(a) – 3, 77 (b) 3, 77
(c) (b – a)/a (d) (a – b)/a
(c) 3, – 77 (d) none of these
18. If  are roots of the equation ax2 + 3x + 2 = 0 (a < 0), then 26. If a, b R, a  b . The roots of the quadratic equation,
 is greater than x2 – 2 (a + b) x + 2 (a2 + b2) = 0 are
(a) 0 (b) 1 (a) Rational and different (b) Rational and equal
(c) 2 (d) none of these
(c) Irrational and different (d) Imaginary and different
19. If  be the roots x2 + px – q = 0 and  be the roots of
27. If the quadratic equations ax 2 + 2cx + b = 0 and
         ax2 + 2bx + c = 0 (b  c) have a common root, then
x2 + px + r = 0 then
         a + 4b + 4c is equal to
(a) –2 (b) –1
(a) 1 (b) q
(c) 0 (d) 1
(c) r (d) q + r
28. If the expression x2 – 11x + a and x2 – 14x + 2a must have a
20. If  be roots of x2 + px + 1 = 0 and  are the roots of
common factor and a  0, then, the common factor is
x2 + qx + 1 = 0 then () () () () =
(a) (x – 3) (b) (x – 6)
(a) p2 + q2 (b) p2 – q2
(c) (x – 8) (d) none of these
(c) q2 – p2 (d) none of these
QUADRATIC EQUATION 23

29. Let a > 0, b > 0 and c > 0. Then both the roots of the 37. The equations x 3 + 5x 2 + px + q = 0 and
equation ax2 + bx + c = 0 x3 + 7x2 + px + r = 0 have two roots in common. If the third
(a) are real and negative (b) have negative real parts root of each equation is represented by x 1 and x 2
respectively, then the ordered pair (x1, x2) is
(c) are rational numbers (d) none of these
(a) (–5, –7) (b) (1, –1)
30. If r and s are positive, then roots of the equation
x2 – rx – s = 0 are (c) (–1, 1) (d) (5, 7)
(a) imaginary 38. If (2x – 3x + 1) (2x + 5x + 1) = 9x2, then equation has
2 2

(b) real and both positive (a) four real roots

(c) real and of opposite signs (b) two real and two imaginary roots
(d) real and both negative (c) all imaginary
31. If a, b, c R and roots of the equation ax2 + 2bx + c = 0 are (d) none of the above
real and different, then roots of the equation 39. If (x + 2) (x + 3) (x + 8) (x + 12) = 4x2, then equation has
(a2 + 2b2 – ac) x2 + 2b (a + c) x + (2b2 + c2 – ac) = 0 are (a) no real roots (b) all real roots
(a) real and equal (b) real and unequal (c) can’t be discussed (d) none of these
(c) imaginary (d) none of these
a
32. If p,q,r,s, R and , are roots of the equation 40. If ‘x’ is real and satisfying the inequality, x   a R  ,
x
x2 + px + q = 0 and 4 and 4 are roots of x2 – rx + s = 0, then
the roots of x2 – 4qx + 2q2 – r = 0 are then

(a) both real (b) both positive (a) x  0,  

a for a  0
(c) both negative (d) none of these
33. If the roots of the quadratic equation x2 – 4x – log3a = 0 are  
(b) x   a , 0 for a  0
real, then the least value of a is
(a) 81 (b) 1/81
 
(c) x   a , 0 for a  0
(c) 1/64 (d) none of these
If  are the roots of the equation, x4 – Kx3 + Kx2 +
34.
Lx + M = 0 where K, L and M are real numbers then the

(d) x   a , 
a for a  0

minimum value of 2 +2 +2 +2 is 41. The set of real ‘x’ satisfying, ||x – 1| –1| < 1 is
(a) 0 (b) –1 (a) [0, 2] (b) [–1, 3]
(c) 1 (d) 2 (c) [–1, 1] (d) [1, 3]
35. The value of ‘a’ for which the sum of the squares of the 42. If one root of the equation 4x2 + 2x – 1 = 0 is , then other
roots of the equation x2 – (a – 2) x – a – 1 = 0 assume the root is
least value is
(a) 2 (b) 4 – 3
(a) 0 (b) 1
(c) 4 + 3 (d) none of these
(c) 2 (d) 3
43. If   are the roots of x2 – p (x + 1) – c = 0 then
36. If the two roots of the equation, x3 – px2 + qx – r = 0 are
equal in magnitude but opposite in sign then  2  2  1 2  2  1
 is equal to
 2  2  c 2  2  c
(a) pr = q (b) qr = p
(c) pq = r (d) none (a) 0 (b) 1
(c) 2 (d) none of these
 24 QUADRATIC EQUATION

44. If the equation (k – 2) x2 – (k – 4) x – 2 = 0 has difference of 53. If b < 0, then the roots x 1 and x 2 of the equation
roots as 3 then the value of k is
x  x 
(a) 1, 3 (b) 3, 3/2 2x2 + 6x + b = 0, satisfy the condition  1    2  < k
 x 2   x1 
(c) 2, 3/2 (d) 3/2, 1
where k is equal to
45. If the equation sin x – (k + 2) sin2 x – (k + 3) = 0 has a
4

solution then k must lie in the interval (a) – 3 (b) – 5

(a) (–4, –2) (b) [–3, 2) (c) – 6 (d) – 2

(c) (–4, –3) (d) [–3, –2] 2x

54. Consider y  , then the range of expression,
46. x 2
The equation,  = –2x + 6x – 9 has 1 x 2
(a) no solution (b) one solution y2 + y – 2 is
(c) two solutions (d) infinite solutions (a) [–1, 1] (b) [0, 1]
2
47. If both roots of the quadratic equation x + x + p = 0 exceed (c) [–9/4, 0] (d) [–9/4, 1]
p where p  R then p must lie in the interval
55. The least value of expression, x2 + 2xy + 2y2 + 4y + 7 is
(a) (–, 1) (b) (–, –2)
(a) –1 (b) 1
(c) (–, –2) (0, 1/4) (d) (–2, 1)
(c) 3 (d) 7
48. If both roots of the quadratic equation (2 – x) (x + 1) = p are 2
56. If the graph of |y| = f (x), where f (x) = ax + bx + c, b, c  R,
distinct and positive then p must lie in the interval
a 0, has the maximum vertical height 4, then
(a) p > 2 (b) 2 < p < 9/4
(a) a > 0 (b) a < 0
(c) p < –2 (d) –  < p <  2
(c) (b – 4ac) is negative (d) Nothing can be said
49. The quadratic equation, ax2 + bx + c = 0 will always have
57. Set of all possible real values of a such that the inequality
imaginary root if 2
(x – (a – 1)) (x – (a + 2)) < 0 holds for all x  (–1, 3) is
(a) a < –1, 0 < c < 1, b > 0
(a) (1, ) (b) (–, –1]
(b) a < –1, –1 < c < 0, 0 < b < 1
(c) (–, –1) (d) (0, 1)
(c) a < –1, c < 0, b > 1
58. If a(p + q) + 2bpq + c = 0 and a(p + r)2 + 2bpr + c = 0,
2

(d) a < –1, c < –1, 1 < b < 2 (a  0) then

50. If b > a and the equation (x – a) (x – b) + 1 = 0 has real roots
2 c
then (a) qr  p  (b) qr = p2
a
(a) both roots in (a, b)
(c) qr = –p2 (d) None of these
(b) both roots in (–, a)
59. If p(x) = ax + bx and q(x) = lx 2 + mx + n with
2
(c) both roots in (b, )
p(1) = q(1); p(2) – q(2) = 1 and p(3) – q(3) = 4, then
(d) one root in (–, a) and other in (b, ) p(4) – q(4) is
51. If  are the roots of the equation, x2 – 2mx + m2 – 1 = 0 (a) 0 (b) 5
then the range of values of m for which (–2, 4) is
(c) 6 (d) 9
(a) (–1, 3) (b) (1, 3)
60. If x  R, then the maximum value of
(c)  ,  1   3,   (d) none

52. If   are the roots of the quadratic equation,


2 2
y = 2(a – x) x + x  b is 
x2 – 2p (x – 4) – 15 = 0 then the set of values of p for which (a) a2 + b2 (b) a2 – b2
one root is less than 1 & the other root is greater than 2 is
(c) a2 + 2b2 (d) none of these
(a) (7/3, ) (b) (–, 7/3)
(c) x  R (d) none
QUADRATIC EQUATION 25

61. If a, b, c R, a > 0 and c 0 Let  and be the real and 68. Let a, b, c  R. If ax2 + bx + c = 0 has two real roots A and
distinct roots of the equation ax2 + bx + c = |c| and p, q be B where A < –1 and B > 1, then
the real and distinct roots of the equation
ax2 + bx + c = 0. Then (a) 1  b  c  0 (b) 1 – b  c  0
a a a a
(a) p and q lie between  and 
(b) p and q do not lie between  and  (c) |c| < |a| (d) |c| < |a| – |b|
(c) Only p lies between  and  2x

(d) Only q lies between  and 

69. 
5x  2 3  – 169  0 is true in the interval.

62. Let f (x) = ax2 + bx + c; a, b, c  R. If f (x) takes real values (a) (–, 2) (b) (0, 2)
for real values of x and non-real values for non-real values (c) (2, ) (d) (0, 4)
of x, then a satisifes.
70. If a < b < c < d, then for any positive , the quadratic
(a) a > 0 (b) a = 0 equation (x – a) (x – c) + (x – b) (x – d) = 0 has
(c) a < 0 (d) a  R (a) non-real roots
3
63. The value of a for which the equations x + ax + 1 = 0 and (b) one real root between a and c
x4 + ax2 + 1 = 0 have a common root is
(c) one real root between b and d
(a) –2 (b) –1
(d) irrational roots
(c) 1 (d) 2
e e   ee
 log x 2 –  9 / 2  log x  5
 3  3
71. Equation    0 has
64. The equation x  
 3 3 has x –e x – x––e

(a) at least one real solution (a) one real root in (e,) and other in ( – e,e)
(b) exactly three real solutions (b) one real root in (e,) and other in (, + e)
(c) exactly one irrational solution (c) two real roots in (– e, + e)
(d) non real roots (d) No real root
65. For a > 0, the roots of the equation 72. If a < 0, then root of the equation x2 – 2a |x – a| – 3a2 = 0 is
logax a + logx a2 + loga2x a3 = 0, are given by

(a) a – 1 – 6  
(b) a 1 – 2 
(a) a–4/3 (b) a–3/4
(c) a–1/2 (d) a–1
2 2 2

(c) a – 1  6  
(d) a 1  2 
66. The roots of the equation, (x + 1) = x(3x + 4x + 3), are
given by 73. If a, b, c  R and  is a real root of the equation
ax2 + bx + c = 0, and  is the real root of the equation
(a) 2 – 3  
(b) – 1  i 3 / 2 , i  –1
a 2
–ax2 + bx + c = 0, then the equation x + bx + c = 0 has
2
(c) 2  3  
(d) – 1 – i 3 / 2 , i  –1
(a) real roots
67. If 0 < a < b < c, and the roots   of the equation (b) none- real roots
ax2 + bx + c = 0 are non real complex roots, then
(c) has a root lying between  and 
(a) || = || (b) || > 1
(d) None of these
(c) || < 1 (d) none of these
 26 QUADRATIC EQUATION

Assertion Reason 78. Assertion : If a > b > c and a3 + b3 + c3 = 3abc, then

(A) If both ASSERTION and REASON are true and the equation ax2 + bx + c = 0 has one
reason is the correct explanation of the assertion. positive and one negative real roots.
(B) If both ASSERTION and REASON are true but reason
Reason : If roots of opposite nature, then
is not the correct explanation of the assertion.
product of roots < 0 and |sums of
(C) If ASSERTION is true but REASON is false.
roots| > 0.
(D) If both ASSERTION and REASON are false.
(E) If ASSERTION is false but REASON is true. (a) A (b) B (c) C (d) D (e) E

74. Assertion : If roots of the equation x2 – bx + c = 0 Using the following passage, solve Q.79 to Q.81
are two consecutive integers, then
b2 – 4c = 1. Passage –1

Reason : If a, b, c are odd integer then the roots In the given figure vertices of  ABC lie on y = f (x)
of the equation 4abc x2 + (b2 – 4ac) = ax2 + bx + c. The ABC is right angled isosceles triangle
x – b = 0 are real and distinct.
whose hypotenuse AC = 4 2 units, then
(a) A (b) B
(c) C (d) D
(e) E
Y
75. Assertion : If one roots is 5 – 2 is then the y = f (x)
= ax2 + bx + c
equation of lowest degree with rational
coefficient is x4 – 14x2 + 9 = 0.
A O C
Reason : For a polynomial equation with X
rational co-efficient irrational roots
B
occurs in pairs.
(a) A (b) B
(c) C (d) D
(e) E 79. y = f (x) is given by
76. Assertion : The set of all real numbers a such that
a2 + 2a, 2a + 3 and a2 + 3a + 8 are the x2 x2
sides of a traiangle is (5, ). (a) y  –2 2 (b) y  –2
2 2 2
Reason : Since in a triangle sum of two sides is
greater than the other and also sides (c) y = x2 – 8
2
(d) y  x – 2 2
are always positive.
(a) A (b) B 80. Minimum value of y = f (x) is
(c) C (d) D
(a) 2 2 (b) – 2 2
(e) E
77. Assertion : The number of roots of the equation (c) 2 (d) – 2
1 x –x
sin (2x) cos (2x) = (2 + 2 ) is 2. k
4 81. Number of integral value of k for which lies between
2
Reason : AM  GM.
the roots of f (x) = 0, is
(a) A (b) B
(c) C (d) D (a) 9 (b) 10
(e) E (c) 11 (d) 12
QUADRATIC EQUATION 27

Using the following passage, solve Q.82 to Q.84 86. Column–I Column–II
Passage –2 (A) Number of real solution of (P) 2
If roots of the equation x4 – 12x3 + bx2 + cx + 81 = 0 are |x + 1| = ex is
positive then (B) The number of non-negative (Q) 3
82. Value of b is real roots of 2x–x–1 = 0 equal to
(a) – 54 (b) 54 (C) If p and q be the roots of the (R) 6
(c) 27 (d) –27
quadratic equation
83. Value of c is
x2 – ( – 2) x –  – 1 = 0, then
(a) 108 (b) –108
minimum value of p2 + q2 is
(c) 54 (d) – 54
equal to
84. Root of equation 2bx + c = 0 is
(D) If  and  are the roots of (S) 5
1 1
(a) – (b) 7
2 2 2x2 + 7x + c = 0 & |2 – 2| = ,
4
(c) 1 (d) –1
then c is equal to
Match the column
Subjective
85. The value of k for which the equation
87. When x100 is divided by x2 – 3x + 2, the remainder is
x3 – 3x + k = 0 has
(2k+1 –1) x – 2(2k – 1) where k is a numerical quantity, then
Column–I Column–II k must be.
(A) three distinct real roots (P) |k| > 2
x
(B) two equal roots (q) k = –2, 2 88. If roots x1 and x2 of x2 + 1 = satisfy
a
(C) exactly one real root (R) |k| < 2
(D) three equal roots (S) no value of k 1 1  1 
x12 – x 22  , then a   – , 0    0 , 
a  2   k

the numerical quantity k must be equal to

89. The integral part of positive value of a for which, the least
value of 4x2 – 4ax + a2 – 2a + 2 on [0, 2] is 3, is
90. If x, y, z are unequal and positive and if x + y + z = 1, the

expression
1 x  1 y  1 z  is greater than
1 – x  1 – y  1 – z 

(The best possible number)

 28 QUADRATIC EQUATION

EXERCISE - 4 : PREVIOUS YEAR JEE ADVANCED QUESTIONS

Single Answer Question
2 2
1. Let a > 0, b > 0 and c > 0. Then, both the roots of the 8. The equation x   1 has (1984)
x 1 x 1
equation ax2 + bx + c = 0 (1979)
(a) no root (b) one root
(a) are real and negative
(c) two equal roots (d) infinitely many roots
(b) have negative real parts
9. If log0.3 (x–1) < log0.09 (x–1), then x lies in the interval
(c) have positive real parts
(1985)
(d) None of the above
(a) (2, ) (b) 1, 2)
2. Both the roots of the equation
(c) (–2, –1) (d) None of these
(x–b) (x–c) + (x–a) (x–c) + (x–a) (x–b) = 0
10. If a, b and c are distinct positive numbers, then the
are always (1980) expression (b + c – a) (c + a – b) (a + b – c) – abc is
(a) positive (b) negative (1986)
(c) real (d) None of these (a) positive (b) negative
3. The least value of the expression 2 log10 x–logx (0.01), (c) non-positive (d) non-negative
for x > 1, is (1980)
11. If and are the roots of x + px + q = 0 and 4, 4 are the
2
(a) 10 (b) 2 roots of x 2 –rx + s = 0, then the equation
(c) –0.01 (d) None of these x2 – 4qx + 2q2 – r = 0 has always, if  and are real numbers.
4. The number of real solutions of the equation (1989)
|x|2 –3|x| + 2 = 0 is (1982) (a) two real roots
(a) 4 (b) 1 (b) two positive roots
(c) 3 (d) 2 (c) two negative roots
5. If x1, x2,....., xn are any real numbers and n is any positive (d) one positive and one negative root
integer, then (1982)
3 5
(log x )2  log x
2 2
2 n n 2 12. The equation x 4 4
 2 has (1989)
n
 n   2 
(a) n  x    x i 2
i (b)  x    x i 
i (a) at least one real solution
i 1  i 1  i 1  i 1 
(b) exactly three real solutions
2
n
2  n
 (c) exactly one irrational solution
(c) x
i 1
i  n   xi 
 i 1 
(d) None of these
(d) complex roots
13. Let a,b,c be real numbers, a  0. If  is a root of
6. The largest interval for which x12 – x9 + x4 – x + 1 > 0 is
a2x2 + bx + c = 0, is the root of a2x2 – bx – c = 0 and
(1982) 0 <  < , then the equation a2x2 + 2bx + 2c = 0 has a root 
(a) – 4 < x 0 (b) 0 < x < 1 that always satisfies (1989)
(c) – 100 < x < 100 (d) –< x < 
  
7. If a + b + c = 0, then the quadratic equation (a)   (b)    
2 2
3ax2 + 2bx + c = 0 has (1983)
(c) =  (d) < < 
(a) at least one root in (0, 1)
14. Let f (x) be a quadratic expression which is positive for all real
(b) one root in (2, 3) and the other in (–2, –1) values of x. If g (x) = f(x) + f’(x) + f’’(x), then for any real x
(c) imaginary roots (1990)
(d) None of the above (a) g (x) < 0 (b) g (x) > 0
(c) g (x) = 0 (d) g (x)  0
QUADRATIC EQUATION 29

15. The number log2 7 is (1990) 24. The number of values of k for which the system of
(a) an integer (b) a rational number equations
(c) an irrational number (d) a prime number (k + 1) x + 8y = 4k
16. Let  be the roots of the equation kx + (k + 3) y = 3k – 1
(x – a) (x – b) = c, c  0 has infinitely many solution, is (2002)
Then the roots of the equation (x – ) (x – ) + c = 0 are (a) 0 (b) 1
(1992) (c) 2 (d) infinite
(a) a, c (b) b, c 25. The set of all real numbers x for which
(c) a, b (d) a + c, b + c x2 – |x + 2| + x > 0 is (2002)

17. The equation x 1  x 1  4x 1 has (1997) (a) (–, –2) (2, ) 

(b)   ,  2    2,  
(a) no solution
(b) one solution (c) (–, –1) (1, ) (d)  2,  
(c) two solutions 26. For all ‘x’, x2 + 2ax + (10 – 3a) > 0, then the interval in which
(d) more than two solutions ‘a’ lies is (2004)
(a) a < –5 (b) –5 < a < 2
 P Q
18. In a triangle PQR, R  , if tan   and tan   are (c) a > 5 (d) 2 < a < 5
2 2 2
27. If one root is square of the other root of the equation
the roots of the equation ax2 + bx + c = 0 (a  0), then
x2 + px + q = 0, then the relation between p and q is
(1999)
(2004)
(a) a + b = c (b) b + c = a 3 2 3 2
(a) p – (3p – 1) q + q = 0 (b) p – q (3p + 1) + q = 0
(c) a + c = b (d) b = c
(c) p3 + q (3p – 1) + q2 = 0 (d) p3 + q (3p + 1) + q2 = 0
19. If the roots of the equation x2 – 2ax + a2 + a – 3 = 0 are real
28. If a, b, c are the sides of a triangle ABC such that
and less than 3, then (1999)
x2 – 2 (a + b + c) x + 3 (ab + bc + ca) = 0 has real roots, then
(a) a < 2 (b) 2 < a < 3
(2006)
(c) 3 < a < 4 (d) a > 4
4 5
20. If  and  ( < ) are the roots of the equation (a)   (b)  
x2 + bx + c = 0, where c < 0 < b, then (2000) 3 3
(a) 0 <  <  (b) < 0 <  < ||  4 5 1 5
(c)    ,  (d)    , 
(c)  < < 0 (d)  < 0 ||<   3 3 3 3
21. If b > a, then the equation (x – a) (x – b) – 1 = 0 has
29. Let  be the roots of the equation x2 – px + r = 0 and
(2000) /2, 2 be the roots of the equation x2 – qx + r = 0. Then the
(a) both roots in (a, b) value of r is (2007)
(b) both roots in (–, a) (a) 2/9 (p – q) (2q – p) (b) 2/9 (q – p) (2p – q)
(c) both roots in (b, + ) (c) 2/9 (q – 2p) (2q – p) (d) 2/9 (2p – q) (2q – p)
(d) one root in (–, a) and the other in (b, + ) 30. Let p and q be the real numbers such that p 0, p3  q and
p3  – q. If  and  are non-zero complex numbers
22. For the equation 3x2 + px + 3 = 0, p > 0, if one of the root is
satisfying + = – p and 3 + 3 = q, then a quadratic
square of the other, then p is equal to (2000)
(a) 1/3 (b) 1  
equation having and as its roots is (2010)
(c) 3 (d) 2/3  

23. The number of solutions of log4 (x–1) = log2 (x – 3) is (a) (p3 + q) x2 – (p3 + 2q) x + (p3 + q) = 0
(2001) (b) (p3 + q) x2 – (p3 – 2q) x + (p3 + q) = 0
(a) 3 (b) 1 (c) (p3 – q) x2 – (5p3 – 2q) x + (p3 – q) = 0
(c) 2 (d) 0 (d) (p3 – q) x2 – (5p3 + 2q) x + (p3 – q) = 0
 30 QUADRATIC EQUATION

31. Let and be the roots of x2 – 6x – 2 = 0, with > . If 36. Let a, b, c, p, q be the real numbers. Suppose , are the
a10  2a 8 1
an = n – n for n 1, then the value of is roots of the equation x2 + 2px + q = 0 and , are the
2a 9 
(2011) roots of the equation ax 2 + 2bx + c = 0, where
2 {–1,0,1}. (2008)
(a) 1 (b) 2
Assertion : (p2 – q) (b2 – ac)  0
(c) 3 (d) 4
Reason : b pa or c qa.
32. A value of b for which the equations x2 + bx – 1 = 0,
x2 + x + b = 0 have one root in common is (2011) (a) (b) (c) (d)
Passage Q. 37–39
(a)  2 (b) i 3 If a continuous f defined on the real line R, assumes positive
and negative values in R, then the equation
(c) i 5 (d) 2 f(x) = 0 has a root in R. For example, if it is known that a
33. The quadratic equation p(x) = 0 with real coefficients has continuous function f on R is positive at some point and
purely imaginary roots. its minimum values is negative, then the equation f(x) = 0
has a root in R.
Then the equation x
Consider f(x) = ke – x for all real x where k is real constant.
p(p(x)) = 0
(2007)
has (2014) x
37. The line y = x meets y = ke for k  0 at
(a) only purely imaginary roots (a) no point (b) one point
(b) all real roots (c) two points (d) more than two points
(c) two real and two purely imaginary roots 38.
x
The positive value of k for which ke – x = 0 has only one
(d) neither real nor purely imaginary roots root is
34. Let S be the set of all non-zero real numbers  such that 1
the quadratic equation x2 – x +  = 0 has two distinct real (a) (b) 1
e
roots x1 and x2 satisfying the inequality |x1 – x2| < 1. Which
of the following intervals is(are) a subset(s) of S ? (c) e (d) loge 2
x
(2015) 39. For k > 0, the set of all values of k for which ke – x = 0 has
two distinct roots, is
 1 1   1 
(a)   ,   (b)   , 0   1 1 
 2 5  5  (a)  0,  (b)  , 1
 e e 
 1   1 1
(c)  0,  (d)  ,  1 
(c)  ,   (d) (0, 1)
 5  5 2 e 
  Passage Q. 40 to 42
35. Let   . Suppose  and  are the roots of the
6  2 3
2
equation x – 2x sec  + 1 = 0 and  and  are the roots Consider the polynomial f (x) = 1 + 2x + 3x + 4x . Let s be
of the equation x2 + 2x tan  1 = 0. If  and , the sum of all distinct real roots of f (x) and let t = |s|
then   equals. (2016) (2010)
(a) 2 (sec  – tan ) (b) 2 sec  40. The real numbers s lies in the interval
(c) 2 tan  (d) 0 3
 1  
Assertion & Reason (a)   ,0  (b)  11,  
 4   4
(A) If ASSERTION is true, REASON is true, REASON is a
correct explanation for ASSERTION.  3 1  1
(c)   ,   (d)  0, 
(B) If ASSERTION is true, REASON is true, REASON is  4 2  4
not a correct explanation for ASSERTION.
(C) If ASSERTION is true, REASON is false
(D) If ASSERTION is false, REASON is true
 QUADRATIC EQUATION 31

41. The area bounded by the curve y = f(x) and the lines True/False
x = 0, y = 0 and x = t, lies in the interval 51. If x – r is a factor of the polynomial
n n–1
3   21 11  f(x) = a n x + a n–1 x + ... + a 0, repeated m times
(a)  ,3  (b)  ,  (1 < m n), then r is a root of f’(x) = 0 repeated m times.
4   64 16 
(1983)
2
 21  52. The equation 2x + 3x + 1 = 0 has an irrational root.
(c) (9, 10) (d)  0, 
 64  (1983)
42. The function f’ (x) is 53. If a < b < c < d, then the roots of the equation
(x–a) (x–c) + 2 (x–b) (x–d) = 0 are real and distinct.
 1  1 
(a) increasing in   t,   and decreasing in   , t  (1984)
 4  4  2 2
54. If P (x) = ax + bx + c and Q(x) = –ax + bx + c, where
ac 0, then P (x) Q (x) has at least two real roots.
 1  1 
(b) decreasing in   t,   and increasing in  ,t (1985)
 4  4 
Subjective Questions
(c) increasing in (–t, t)
(d) decreasing in (–t, t) 26  15 3
55. Show that the square of is a rational
Passage Q. 43 and 44
5 2  (38  5 3)
Let p, q be integers and let ,  be the roots of the
number. (1978)
equation, x2 – x – 1 = 0, where   . For n = 0, 1, 2,……., 2
56. If and are the roots of the equation x + px + 1 = 0;
let an = pn + qn. 2
,  are the roots of x + qx + 1 = 0, then
2 2
FACT : If a and b are rational numbers and a  b 5  0, q – p = (– ) (– ) (+ ) (+ ) (1978)
then a = 0 = b. (2017) 57. Solve 2 logx a + logax a + 3 logb a = 0,
2
43. If a4 = 28, then p + 2q = where a > 0, b = a x (1978)
2
(a) 12 (b) 21 58. If and are the roots of x + px + q = 0 and , are the
2
roots of x + rx + s = 0, then evaluate
(c) 14 (d) 7
(– ) (– ) (– ) (– ) in terms of p, q, r and s.
44. a12 =
(1979)
(a) a11 + 2a10 (b) a11 + a10
59. Show that for any triangle with sides a,b,c;
2
(c) a11 – a10 (d) 2a11 + a10 3(ab + bc + ca) (a + b + c) 4 (ab + bc + ca) (1979)
Fill in the Blanks 60. Find the integral solutions of the following systems of
2
inequalities
45. If 2  i 3 is a root of the equation x + px + q = 0, where p 2
(a) 5x – 1 < (x + 1) < 7x – 3
and q are real, then (p, q) = (.......). (1982)
46. If the products of the roots of the equation x 1 6x 1
2 2 log k (b)  ,  (1979)
x –3kx + 2e – 1 = 0 is 7, then the roots are real for 2x  1 4 4x  1 2
k = ... . (1984)
2
61. For what values of m, does the system of equations
47. If the quadratic equations x + ax + b = 0 and
2 3x + my = m
x + bx + a = 0 (a  b) have a common root, then the numerical
value of a + b is ... . (1986) and 2x – 5y = 20
has solution satisfying the conditions x > 0, y > 0 ?
48. The solution of the equation log7 log5 ( x  5  x )  0
(1980)
is .... (1986) 2
62. If one root of the quadratic equation ax + bx + c = 0 is
49. If , , are the cube roots of p, p < 0, then for any x, y and equal to the nth power of the other, then show that
x  y  z
z, then  ... (1990) 1 1
x  y  z (ac n ) n 1  (a n c) n 1  b  0 (1983)

50. The sum of all the real roots of the equation

2
|x–2| + |x–2| – 2 = 0 is .... . (1997)
 32 QUADRATIC EQUATION
QUADRATIC EQUATION
2 2
63. Find all real values of x which satisfy x – 3x + 2 > 0 and x 75. Let –1 < p < 1. Show that the equation 4x3 – 3x – p = 0 has
– 2x – 4  0. (1983) a unique root in the interval [1/2, 1] and identify it.
64. If a > 0, b > 0 and c > 0 prove that (2001)
76. Let a, b, c be real numbers with a  0 and let  be the
1 1 1 roots of the equation ax2 + bx + c = 0. Express the roots of
(a  b  c)      9 (1984)
a b c a3x2 + abcx + c3 = 0 in terms of , . (2001)
77. If x2 + (a – b) x + (1 – a – b) = 0 where a, b  R, then find the
x2 3 x 2 3
65. Solve for x (5  2 6)  (5  2 6)  10 (1985) values of a for which equation has unequal real roots for
all values of b. (2003)
66. For a  0, determine all real roots of the equation
2 2 78. If  are the roots of ax2 + bx + c = 0, (a 0) and + ,
x – 2a |x–a| – 3a = 0 (1986)
2
+ , are the roots of Ax2 + Bx + C = 0, (A 0) for some
67. Solve |x + 4x + 3| + 2x + 5 = 0 (1987)
b2  4ac B2  4AC
68. Find the set of all x for which constants , then prove that  (2004)
a2 A2
2x 1
2
 (1987) 79. If x 2 – 10ax – 11b = 0 have roots c & d,
2x  5x  2 x  1 x2 – 10cx – 11d = 0 have roots a and b. (a  c) Find
69. Solve the x the following equation a + b + c + d. (2006)
2
log(2x + 3) (6x + 23x + 21) Integer Answer Type Questions
2
= 4 – log(3x + 7) (4x + 12x + 9) (1987) 80. The smallest value of k, for which both the roots of the
2 2
2 equation x – 8kx + 16 (k – k + 1) = 0 are real, distinct and
70. Let a, b, c be real. If ax + bx + c = 0 has two real roots and
have values at least 4, is.... (2009)
, where < – 1 and > 1, then show that
81. Let (x, y, z) be points with integer coordinates satisfying
c b the system of homogeneous equations
1  0 (1995)
a a 3x – y – z = 0, –3x + z = 0, –3x + 2y + z = 0. Then the number
2 2 2
of such points for which x + y + z  100 is...
71. Let S be a square of unit area. Consider any quadrilateral
which has one vertex on each side of S. If a, b, c and d (2009)
denote the length of the sides of the quadrilateral, prove 2 2
that 2 < a2 + b2 + c2 + d2 < 4 (1997) 82. Let be the complex number cos
 i sin . Then the
3 3
72. Find the set of all solutions of the equation number of distinct complex number z satisfying
2|y| – |2y – 1 – 1| = 2y – 1 + 1 (1997)
z 1  2
73. Let f (x) = Ax2 + Bx + C where, A, B, C are real numbers.
Prove that if f (x) is an integer whenever x is an integer,  z  2 1  0 is equal to..... (2010)
then the numbers 2A, A + B and C are all integers. 2 1 z
Conversely, prove that if the numbers 2A, A + B and C are
all integers, then f (x) is an integer whenever x is an integer. 83. Let a, b, c be three non-zero real numbers such that the
(1998)
  
74.
2
If ,  are the roots of ax + bx + c = 0, (a  0) and equation 3acosx  2bsin x  c, x    ,  , has two
2  2 2
+ , + are the roots of Ax + Bx + C = 0, (A 0) for some
constant , then prove that 
distinct real roots  and  with   . . Then, the value
3
b 2  4ac B2  4AC
 (2000)
a2 A2 b
of is _________. (2018)
a
 QUADRATIC EQUATION 33

ANSWER KEY
EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS
1. (b) 2. (c) 3. (d) 4. (d) 5. (c) 6. (d) 7. (b) 8. (b) 9. (d) 10. (a)
11. (a) 12. (c) 13. (a) 14. (a) 15. (b) 16. (b) 17. (b) 18. (a) 19. (c) 20. (d)
21. (b) 22. (a) 23. (d) 24. (c) 25. (b) 26. (c) 27. (c) 28. (d) 29. (a) 30. (b)
31. (b) 32. (c) 33. (b) 34. (d) 35. (b) 36. (b) 37. (b) 38. (a) 39. (b) 40. (a)
41. (b) 42. (d) 43. (a) 44. (a) 45. (a,b) 46. (c) 47. (c) 48. (a) 49. (b) 50. (b,c)
51. (c) 52. (a) 53. (a) 54. (b) 55. (b) 56. (b) 57. (b) 58. (a) 59. (a) 60. (a)
61. (a) 62. (d) 63. (a) 64. (a,b,c,d) 65. (b) 66. (a,b,c,d) 67. (b) 68. (a) 69. (d) 70. (b)
71. (c) 72. (b) 73. (c) 74. (d) 75. (d) 76. (c) 77. (b) 78. (d) 79. (d) 80. (c)

EXERCISE - 2 : PREVIOUS YEAR JEE MAINS QUESTIONS

1. (d) 2. (a) 3. (c) 4. (a) 5. (a) 6. (d) 7. (a) 8. (b) 9. (a) 10. (c)
11. (b) 12. (c) 13. (d) 14. (b) 15. (c) 16. (c) 17. (a) 18. (c) 19. (c) 20. (b)
21. (b) 22. (a) 23. (b) 24. (b) 25. (a) 26. (b) 27. (c) 28. (a) 29. (a) 30. (c)
31. (c) 32. (d) 33. (d) 34. (a) 35. (d) 36. (c) 37. (b) 38. (c) 39. (b) 40. (d)
41. (b)

EXERCISE - 3 : ADVANCED OBJECTIVE QUESTIONS

1. (d) 2. (a,c) 3. (a) 4. (c) 5. (c) 6. (d) 7. (d) 8. (a) 9. (a) 10. (a)
11. (c) 12. (b) 13. (c) 14. (b) 15. (a) 16. (a) 17. (a) 18. (d) 19. (a) 20. (c)
21. (d) 22. (c) 23. (a) 24. (c) 25. (a,d) 26. (d) 27. (c) 28. (c) 29. (b) 30. (c)
31. (c) 32. (d) 33. (b) 34. (b) 35. (b) 36. (c) 37. (a) 38. (a) 39. (b) 40. (a,c)
41. (b) 42. (b) 43. (b) 44. (b) 45. (d) 46. (a) 47. (b) 48. (b) 49. (d) 50. (a)
51. (a) 52. (b) 53. (d) 54. (c) 55. (c) 56. (b) 57. (b) 58. (a) 59. (d) 60. (a)
61. (a) 62. (b) 63. (a) 64. (a,b,c) 65. (a,c) 66. (a,b,c,d) 67. (a,b) 68. (a,b) 69. (a,b) 70. (b,c)
71. (b,c) 72. (b,c) 73. (a,c) 74. (b) 75. (a) 76. (a) 77. (e) 78. (a) 79. (a) 80. (b)
81. (c) 82. (b) 83. (b) 84. (c) 85. A–R; B–Q; C–P; D–S 86. A–Q; B–P; C–S; D–R
87. 0099 88. 0005 89. 0008 90. 0008

EXERCISE - 4 : PREVIOUS YEAR JEE ADVANCED QUESTIONS

1. (b) 2. (c) 3. (d) 4. (a) 5. (d) 6. (d) 7. (a) 8. (a) 9. (a) 10. (b)
11. (a,d) 12. (a,b,c) 13. (d) 14. (b) 15. (c) 16. (c) 17. (a) 18. (a) 19. (a) 20. (b)
21. (d) 22. (c) 23. (b) 24. (b) 25. (b) 26. (b) 27. (a) 28. (a) 29. (d) 30. (b)
31. (c) 32. (b) 33. (d) 34. (a, d) 35. (c) 36. (b) 37. (b) 38. (a) 39. (a) 40. (c)
41. (a) 42. (b) 43. (a) 44. (b) 45. (–4, 7) 46. k = 2 47. –1 48. x = 4 49. w2 50. 4
51. False 52. False 53. True 54. True 57. x = a–1/2 or a–4/3 58. (q – s)2 – rqp – rsp + sp2 + qr2
 15 
60. (a) x = 3 (b) x =  61. m   ,    (30, ) 63. x  [1  5, 1)  [1  5, 2) 65. x   2,  2
 2

66. x  {a (1  2), a ( 6  1)} 67. x  4, (1  3)

 2 1
68. x  (2, 1)    ,  
 3 2
1 1   
69. x   72. y {–1} [1, ) 75. x  cos  cos 1 p  76. x =  
4  3 
77. a > 1 79. 1210 80. k = 2 81. 7 82. 1 83. (0.5)