Quadratic

Equations 4

BASIC CONCEPTS – A FLOW CHART

Quadratic Equations 77
 MORE POINTS TO REMEMBER
2
 If ax + bx + c is factorizable into a product of two linear factors, then the roots of the
quadratic equation ax2 + bx + c = 0 can be found by equating each factors to zero.
For example, x2 – 5x + 6 = 0 ⇒ x2 – 3x – 2x + 6 = 0
⇒ x(x – 3) – 2 (x – 3) = 0 ⇒ (x – 3) (x – 2) = 0
⇒ x – 3 = 0 or x – 2 = 0 ⇒ x = 3 or x = 2
2
⇒ x = 3, 2 are roots of x – 5x + 6 = 0.
 Sridharacharya’s Formula: The following formula used to find the roots a, b of quadratic
2 – b + b2 – 4ac – b – b2 – 4ac
equation ax + bx + c = 0, a ≠ 0, is given by a = and b = are
2a 2a
called Sridharachaya’s formula as it was first given by an ancient Indian Mathematician
Sridharacharya in 1025 AD.

Multiple Choice Questions [1 mark]

Choose and write the correct option in the following questions.
1. Which of the following is a quadratic equation? [NCERT Exemplar]
2
2
(a) x + 2x + 1 = (4 – x) + 3 2
(b) – 2x = (5 – x) d 2x – n
2
5
3
(c) (k + 1)x2+ x = 7 , where k = – 1 (d) x3 – x2 = (x – 1)3
2
2. Which of the following is not a quadratic equation? [NCERT Exemplar]
(a) 2(x – 1)2 = 4x2 – 2x + 1 (b) 2x – x2 = x2 + 5
(c) ^ 2 x + 3 h + x2 = 3x2 – 5x (d) (x2 + 2x)2 = x4 + 3 + 4x3
2

3. Which of the following equations has 2 as a root? [NCERT Exemplar]

2 2
(a) x – 4x + 5 = 0 (b) x + 3x – 12 = 0
(c) 2x2 – 7x + 6 = 0 (d) 3x2 – 6x – 2 = 0
4. Which of the following equations has the sum of its roots as 3? [NCERT Exemplar]
(a) 2x2 – 3x + 6 = 0 (b) – x2 + 3x – 3 = 0
3
(c) 2 x2 – x + 1 = 0 (d) 3x2 – 3x + 3 = 0
2
5. Value(s) of k for which the quadratic equation 2x2 – kx + k = 0 has equal roots is
 [NCERT Exemplar]
(a) 0 (b) 4 (c) 8 (d) 0 and 8
6. The quadratic equation 2x2 – 5 x + 1 = 0 has  [NCERT Exemplar]
(a) two distinct real roots (b) two equal real roots
(c) no real root (d) more than two real roots
7. The two consecutive odd positive integers, sum of whose squares is 290 are
(a) 13, 15 (b) 11, 13 (c) 7, 9 (d) 5, 7
8. (x2 + 1)2 – x2 = 0 has
(a) four real roots (b) two real roots
(c) no real roots (d) one real root

78 Xam idea Mathematics–X

 5+ 5
9. The quadratic equation with real co-efficients whose one root is , is
2
(a) x2 – 5x + 5 = 0 (b) x2 + 5x – 5 = 0
(c) x2 – 5x – 5 = 0 (d) x2 – 5x + 1 = 0
10. Root of the equation x2 – 0.09 = 0 is
(a) 0.3 (b) 0.03 (c) no root (d) none of these
11. If ax2 + bx + c = 0 has equal roots, then the value of b is
(a) ! ac (b) ! 2 ac (c) ac (d) none of these
1 5
is a root of the equation x2 + kx – = 0, then the value of k is
12. If
2 4
1 1
(a) 2 (b) – 2 (c) (d)
4 2
13. If the list price of a toy is reduced by ™ 2, a person can buy 2 toys more for ™ 360. The original
price of the toy is
(a) ™18 (b) ™ 20 (c) ™ 19 (d) ™ 21
14. If the equation x2 – kx +9 = 0 does not possess real roots, then
(a) – 6 < k < 6 (b) k > 6 (c) k < – 6 (d) k = ± 6
15. Which of the following is a quadratic equation?
(a) (4 – x) (3x + 1) = 2 – 3x2 (b) (x + 3) 2 – 5 = x2 + 9
(c) (k – 4) x2 – 3x = 9k – 4 , (k = 4) (d) (x + 1)3 = x3 – 5
16. Which of the following equations has – 1 as a root?
(a) x2 + 3x – 10 = 0 (b) x2 – x – 12 = 0
(c) 3x2 – 2x – 5 = 0 (d) 9x2 + 24x + 16 = 0
17. If the difference of roots of the quadratic equation x2 + kx + 12 = 0 is 1, the positive value of k is
(a) – 7 (b) 7 (c) 4 (d) 8
18. The quadratic equation whose one rational root is 3 + 2 is
(a) x2 – 7x + 5 (b) x2 + 7x + 6 = 0
2
(c) x – 7x + 6 (d) x2 – 6x + 7 = 0

19. Which of the following equations has two distinct real roots? [NCERT Exemplar]
9
(a) 2x2 – 3 2 x + = 0 (b) x2 + x – 5 = 0
4
(c) x2 + 3x + 2 2 = 0 (d) 5x2 – 3x + 1 = 0
20. Which of the following equations has no real roots? [NCERT Exemplar]
(a) x2 – 4x + 3 2 = 0 (b) x2 + 4x – 3 2 = 0
(c) x2 – 4x – 3 2 = 0 (d) 3x2 + 4 3 x + 4 = 0

Answers
1. (d) 2. (c) 3. (c) 4. (b) 5. (d) 6. (c)
7. (b) 8. (c) 9. (a) 10. (a) 11. (b) 12. (a)
13. (b) 14. (a) 15. (d) 16. (c) 17. (b) 18. (d)
19. (b) 20. (a)

Quadratic Equations 79
Fill in the Blanks [1 mark]
Complete the following statements with appropriate word(s) in the blank space(s).
A real number α is said to be a _______________ of the quadratic equation ax2 + bx + c = 0, if
1.
aα2 + bα + c = 0.
For any quadratic equation ax2 + bx + c = 0, b2 – 4ac, is called the _______________ of the
2.
equation.
3.
A polynomial of degree 2 is called the _______________ polynomial.
4.
If the discriminant of a quadratic equation is zero, then its roots are _______________ and
_______________ .
5.
If the discriminant of a quadratic equation is greater than zero, then its roots are
_______________and _______________ .
6.
A quadratic equation does not have any real roots if the value of its discriminant is
_______________ zero.
7.
The roots of a quadratic equation is same as the _______________ of the corresponding quadratic
polynomial.
8. If the equation x2 + x – 5 = 0 then product of its two roots is _______________ .
The equation of the form ax2 + bx = 0 will always have _______________ roots.
9.
10. If the product ac in the quadratic equation ax2 + bx + c is negative, then the equation cannot
have _______________ roots.
11. _______________ is a root of quadratic equation x2 – 0.04 = 0.
2
12. If one root of quadratic equation 6x2 – x – k = 0 is , value of k is _______________.
3
– b + b2 – 4ac
13. If one root of a quadratic equation is , other root is _______________.
2a
14. If the coefficient of x2 and constant term of a quadratic equation have _______________ signs
then the quadratic equation has real roots.
15. If 2 is a zero of the quadratic polynomial p(x) then 2 is a root of the quadratic equation
_______________.

Answers
1. root 2. discriminant 3. quadratic 4. real, equal 5. real, distinct 6. less than
7. zeros 8. – 5 9. real 10. non-real 11.  0.2 12. 2
2
–b – b – 4ac
13. 14. opposite 15. p(x) = 0
2a

Very Short Answer Questions [1 mark]

1. What will be the nature of roots of quadratic equation 2x2 – 4x + 3 = 0? [CBSE 2019 (30/2/1)]
Sol. D = b2 – 4ac = 42 – 4 × 2 × 3 = 16 – 24 = –8 < 0
Since D < 0
Hence, roots are not real.
2. If x = 3 is one root of the quadratic equation x2 – 2kx – 6= 0, then find the value of k.
[CBSE 2018]
2
Sol.  x = 3 is a root of the equation x – 2k x – 6 = 0.
 32 – 2k (3) – 6 = 0 ⇒ 9 – 6k – 6 = 0
1
⇒ 3 – 6k = 0 ⇒ 3 = 6k ⇒ k = .
2
80 Xam idea Mathematics–X
 3. Find the value of k for which the quadratic equation kx (x – 2) + 6 = 0 has two equal roots.
 [CBSE 2019 (30/4/1)]
2
Sol. Given equation kx –2kx + 6 = 0
For two equal roots
D=0
2
⇒ b – 4ac = 0
⇒ 4k2 – 4k × 6 = 0
⇒ 4k (k – 6) = 0  ⇒  k = 6 [k ≠ 0, as if k = 0 then the given equation is not a valid equation.]

4. If a and b are the roots of the equation x2 + ax – b = 0, then find a and b.
B
Sol. Sum of the roots = a + b = –
=– a
A
C
Product of the roots = ab = = – b
A
⇒ a + b = – a and ab = – b
⇒ 2a = – b
and a = –1 ⇒ b = 2 and a = – 1
2
5. Show that x = – 2 is a solution of 3x + 13x + 14 = 0.
Sol. Put the value of x in the quadratic equation,
LHS = 3x2 + 13x + 14 = 3(–2)2 + 13(–2) + 14
= 12 – 26 + 14 = 0 = RHS
Hence, x = – 2 is a solution.
6. Write the discriminant of the quadratic equation: (x + 5)2 = 2(5x – 3).
[CBSE 2019 (30/3/1), (30/3/3)]
Sol. Given equation:
(x + 5)2 = 2(5x – 3)
x2 + 25 + 10x = 10x – 6
⇒
x2 + 31 = 0
⇒
Here  a = 1, b = 0, c = 31
D = b2 – 4ac
  = – 4 × 1 × 31 = –124
∴
Discriminant = –124

Short Answer Questions-I [2 marks]

1. State whether the equation (x + 1)(x – 2) + x = 0 has two distinct real roots or not. Justify your
answer.
Sol. (x + 1)(x – 2) + x = 0 ⇒ x2 – x – 2 + x = 0   ⇒  x2 – 2 = 0
D = b2 – 4ac = 0 – 4(1) (–2) = 8 > 0
\ Given equation has two distinct real roots.

2. Is 0.3 a root of the equation x2 – 0.9 = 0? Justify.
Sol. If 0.3 is a root of the equation x2 – 0.9 = 0, then
x2 – 0.9 = (0.3)2 – 0.9 = 0.09 – 0.9 ≠ 0

Hence, 0.3 is not a root of given equation.

Quadratic Equations 81
 3. Find the value of k, for which x = 2 is a solution of the equation kx2 + 2x – 3 = 0.
 [CBSE 2019 (30/5/1)]
2
Sol. If x = 2 is a solution of kx + 2x – 3 = 0, then
k(22) + 2(2) – 3 = 0

⇒ 4k + 4 – 3 = 0

–1
⇒ 4k = –1
⇒k=
4
4. Find the value of k for which the equation x2 + k(2x + k − 1)+ 2 = 0 has real and equal
roots.  [CBSE Delhi 2017]
Sol. Given quadratic equation: x2 + k(2x + k – 1) + 2 = 0
⇒
x2 + 2kx + (k2 – k + 2) = 0
For equal roots, b2 – 4ac = 0
⇒ 4k2 – 4k2 + 4k – 8 = 0

⇒ 4k = 8
⇒ k = 2
5. If –5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation
p(x2 + x) + k = 0 has equal roots, then find the value of k. [CBSE (F) 2014, (AI) 2016]
Sol. Since – 5 is a root of the equation 2x2 + px – 15 = 0
\ 2(–5)2 + p(–5) – 15 = 0

⇒
50 – 5p – 15 = 0 or 5p = 35 or p=7
2 2
Again p(x + x) + k = 0 or 7x + 7x + k = 0 has equal roots.
\
D=0
2 49 7

i.e., b – 4ac = 0 or 49 – 4 × 7k = 0 ⇒ = k=
28 4
6. Does there exist a quadratic equation whose co-efficients are rational but both of its roots are
irrational? Justify your answer.
Sol. Yes, x2 – 4x + 1 = 0 is a quadratic equation with rational co-efficients.
4 ! (– 4) 2 – 4 # 1 # 1 4 ! 12
Its roots are = = 2 ! 3 , which are irrational.
2 2
7. Solve the quadratic equation 2x2 + ax – a2 = 0 for x. [CBSE Delhi 2014]
2 2
Sol. 2x + ax – a = 0
Here, a = 2, b = a and c = – a2
Using the formula,
– b ! b2 – 4ac
x= , we get
2a

– a ! a 2 – 4 # 2 # ( – a 2 ) – a ! 9a 2 – a ! 3 a
x= = =
2#2 4 4
– a + 3a a – a – 3a a
x= = , x= =– a ⇒ x= , – a
4 2 4 2

8. Find the roots of the quadratic equation 2 x2 + 7x + 5 2 = 0 . [CBSE Delhi 2017]

Sol. The given quadratic equation is
2 x 2 + 7x + 5 2 = 0
By applying mid term splitting, we get
2 x 2 + 2x + 5x + 5 2 =0

82 Xam idea Mathematics–X

 ⇒
2 x ( x + 2 ) + 5 ( x + 2 ) = 0 ⇒ ( 2 x + 5 ) (x + 2 ) = 0
–5 –5 2
⇒
x= , – 2 or ,– 2
2 2

2
9. If x = and x = –3 are roots of the quadratic equation ax2 + 7x + b = 0, find the values of a
3
and b. [CBSE Delhi 2016]
Sol. Let us assume the quadratic equation be Ax2 + Bx + C = 0.
B
Sum of the roots = –
A
–7 2
⇒ a = 3 –3 & a = 3
C
Product of the roots =
A
b 2 b
⇒
= × (−3) ⇒ = −2 ⇒ b = 3 × (−2) ⇒ b = −6
a 3 a

Short Answer Questions-II [3 marks]

1. A two digit number is four times the sum of the digits. It is also equal to 3 times the product
of digits. Find the number. [CBSE (F) 2016]
Sol. Let the ten’s digit be x and unit’s digit = y
Number = 10x + y
\
10x + y = 4(x + y) ⇒ 6x = 3y ⇒   2x = y ...(i)
Again 10x + y = 3xy
10x + 2x = 3x (2x) ⇒   12x = 6x2 [From equation (i)]
⇒
6x2 – 12x = 0 ⇒ 6x(x – 2) = 0 ⇒    x = 2 (rejecting x = 0)
From (i), 2x = y ⇒ y = 4
\ The required number is 24.

2. Find the value of p, for which one root of the quadratic equation px2 – 14x + 8 = 0 is 6 times
the other. [CBSE (AI) 2017]
Sol. Let the roots of the given equation be a and 6a.
Thus the quadratic equation is (x – a) (x – 6a) = 0
⇒
x2 – 7ax + 6a2 = 0 ...(i)
Given equation can be written as
14 8
x2 – x + = 0 ...(ii)
p p
14 8
Comparing the co-efficients in (i) & (ii) 7α = and 6α2 =
p p
14 2
7α = ⇒ α=
p p
2
8 2 8
6×e o = ⇒ p =
6×4
and 6α2 = ⇒ ⇒p=3
p p p 8
3. If ad ≠ bc, then prove that the equation
(a2 + b2) x2 + 2(ac + bd)x + (c2 + d2) = 0 has no real roots. [CBSE (AI) 2017]
2 2 2 2 2
Sol. The given quadratic equation is (a + b )x + 2(ac + bd)x + (c + d ) = 0.
D = b2 – 4ac

Quadratic Equations 83
 = 4(ac + bd)2 – 4(a2 + b2) (c2 + d2)
= – 4(a2d2 + b2c2 – 2abcd) = – 4(ad – bc)2
Since ad ≠ bc and (ad – bc)2 > 0
Therefore D < 0
Hence, the equation has no real roots.

4. Find the roots of the following quadratic equation by factorisation:
1
2x2 – x + = 0
8  [NCERT]
1
Sol. We have, 2x2 − x + =0
8
16x2 – 8x +1
⇒ =0
8
⇒ 16x2 – 8x + 1 = 0

⇒ 16x2 – 4x – 4x + 1 = 0

⇒ 4x (4x – 1) – 1(4x – 1) = 0

⇒ (4x – 1) (4x – 1) = 0
So, either 4x – 1 = 0 or 4x – 1 = 0
1 1
x= or x =
4 4
1 1
Hence, the roots of the given equation are and .
4 4
5. Find the roots of the following quadratic equation by applying the quadratic formula.
4x2 + 4 3 x + 3 = 0 [NCERT]
2
Sol. We have, 4x + 4 3 x + 3 = 0
Here, a = 4, b = 4 3 and c = 3
Therefore, D = b2 – 4ac = (4 3 ) 2 – 4 × 4 × 3 = 48 – 48 = 0
 D = 0, roots exist and are equal.

– b! D –4 3 ! 0 – 3
Thus, x= = =
2a 2 ×4 2
– 3 – 3
Hence, the roots of given equation are and .
2 2

6. Using quadratic formula solve the following quadratic equation:

p2x2 + (p2 – q2) x – q2 = 0

Sol. We have, p2x2 + (p2 – q2) x – q2 = 0
Comparing this equation with ax2 + bx + c = 0, we have
a = p2, b = p2 – q2 and c = – q2
\ D = b2 – 4ac = (p2 – q2)2 – 4 × p2 × (– q2)

= (p2 – q2)2 + 4p2q2 = (p2 + q2)2 > 0
So, the given equation has real roots given by

84 Xam idea Mathematics–X

 2 2 2 2 2
– b + D – ( p – q ) + ( p + q ) 2q q2
a= = = 2= 2
2a 2p 2 2p p
2 2 2 2 2
– b – D – ( p – q ) – ( p + q ) – 2p
and b= = = = –1
2a 2p2 2p2
q2
Hence, roots are 2 and –1.
p
7. Find the nature of the roots of the following quadratic equation. If the real roots exist, find them:
3x2 – 4 3 x + 4 = 0
[NCERT]
Sol. We have, 3x2 – 4 3 x + 4 = 0
Here, a = 3, b = – 4 3 and c = 4
Therefore, D = b2 – 4ac = (– 4 3 ) 2 – 4 × 3 × 4 = 48 – 48 = 0
Hence, the given quadratic equation has real and equal roots.
– b ! D – (– 4 3 ) ! 0 2 3
Thus, x= = =
2a 2#3 3

2 3 2 3
Hence, equal roots of given equation are , .
3 3
8. Find the value of k for the following quadratic equation, so that they have two equal roots.
kx (x – 2) + 6 = 0
[NCERT]
Sol. We have, kx(x – 2) + 6 = 0 ⇒ kx2 – 2kx + 6 = 0
Here, a = k, b = – 2k, c = 6
For equal roots, we have
D=0
2

i.e., b – 4ac = 0 ⇒ (–2k)2 – 4 × k × 6 = 0
⇒ 4k2 – 24k = 0
⇒ 4k (k – 6) = 0
Either 4k = 0 or k – 6 = 0 ⇒ k = 0 or k = 6
But k ≠ 0 (because if k = 0 then given equation will not be a quadratic equation).
So, k = 6.
9. If the roots of the quadratic equation (a – b) x2 + (b – c) x + (c – a) = 0 are equal, prove that
2a = b + c. [CBSE (F) 2017]
Sol. Since the equation (a – b) x2 + (b – c) x + (c – a) = 0 has equal roots, therefore discriminant
D = (b – c)2 – 4(a – b) (c – a) = 0
= b2 + c2 – 2bc – 4(ac – a2 – bc + ab) = 0
= b2 + c2 – 2bc – 4ac + 4a2 + 4bc – 4ab = 0
= 4a2 + b2 + c2 – 4ab + 2bc – 4ac = 0
= (2a)2 + (– b)2 + (– c)2 + 2(2a) (– b) + 2(– b) (– c) + 2 (– c) 2a = 0
= (2a – b – c)2 = 0 ⇒ 2a – b – c = 0 ⇒ 2a = b + c.
Hence proved
10. If the equation (1 + m2) x2 + 2mcx + c2 – a2 = 0 has equal roots, show that c2 = a2 (1 + m2).
 [CBSE Delhi 2017]
Sol. The given equation is (1 + m2) x2 + (2mc) x + (c2 – a2) = 0

Quadratic Equations 85
 Here, A = 1 + m2, B = 2mc and C = c2 – a2
Since the given equation has equal roots, therefore D = 0 ⇒ B2 – 4AC = 0.
⇒ (2mc)2 – 4(l + m2) (c2 – a2) = 0

⇒ 4m2c2 – 4(c2 – a2 + m2c2 – m2a2) = 0

⇒
m2c2 – c2 + a2 – m2c2 + m2a2 = 0 [Dividing throughout by 4]
2 2 2 2 2 2
⇒ – c + a (1 + m ) = 0
⇒ c = a (1 + m )
Hence proved
11. If sin q and cos q are roots of the equation ax2 + bx + c = 0, prove that a2 – b2 + 2ac = 0.
−B –b
Sol. Sum of the roots = ⇒ sin q + cos q = a ...(i)
A
C c
Product of the roots = A ⇒ sin q . cos q = a ...(ii)

Now, we have, sin2 q + cos2 q = 1 2

–b c
c
⇒
(sin q + cos q)2 – 2sin q cos q = 1 ⇒ a m – 2 . a =1

b 2 2c
⇒
– a =1    or b2 – 2ac = a2
a2
⇒
a2 – b2 + 2ac = 0
12. Determine the condition for one root of the quadratic equation ax2 + bx + c = 0 to be thrice
the other.
Sol. Let the roots of the given equation be a and 3a.
–b
Then sum of the roots = a + 3a = 4a = a ...(i)
c
Product of the roots = (a)(3a) = 3a2 = a ...(ii)
–b
From (i), a =
4a
–b 2 c 3b 2
(ii) ⇒ 3 c m =a
c
⇒ =a
4a 16a 2

⇒ 3b2 = 16ac, which is the required condition.

2x – 1 x+3
13. Solve for x: 2 c m – 3c m = 5; x ! – 3,
1
[CBSE (F) 2014]
x+3 2x – 1 2

x+3 4x – 2 3x + 9
c m–c m=5
2x – 1
Sol. 2c m – 3c m = 5 ⇒
x+3 2x – 1 +
x 3 2x – 1

(4x – 2)(2x – 1) – (3x + 9)(x + 3) = 5(x + 3)(2x – 1)

(8x2 – 4x – 4x + 2) – (3x2 + 9x + 9x + 27) = 5(2x2 – x + 6x – 3)
8x2 – 8x + 2 – 3x2 – 18x – 27 = 10x2 + 25x – 15
5x2 – 26x – 25 = 10x2 + 25x – 15
5x2 + 51x + 10 = 0
5x2 + 50x + x + 10 = 0
5x (x + 10) + 1 (x + 10) = 0
(5x + 1) (x + 10) = 0
5x + 1 = 0 or x + 10 = 0
–1
x = or x = – 10
5
86 Xam idea Mathematics–X
 14. Solve for x: x2 + 5x – (a2 + a – 6) = 0 [CBSE (F) 2015]
2 2
Sol. x + 5x – (a + a – 6) = 0
x + 5x – (a2 + 3a – 2a – 6) = 0
2

x2 + 5x – [a(a + 3) –2 (a + 3)] = 0
x2 + 5x – (a – 2) (a + 3) = 0
\
x2 + (a + 3)x – (a – 2) x – (a – 2) (a + 3) = 0
x[x + (a + 3)] – (a – 2) [x + (a + 3)] = 0
[{x + (a + 3)} {x – (a – 2)}] = 0
\
x = – (a + 3) or x = (a – 2) ⇒
x = – (a + 3), (a – 2)
Alternative method

x2 + 5x – (a2 + a – 6) = 0
– 5 ! 52 – 4 # 1 # [– (a2 + a – 6)]
\
x=
2#1

– 5 ! 25 + 4a2 + 4a – 24 – 5 ! 4a 2 + 4 a + 1
= =
2 2
– 5 ! ^2ah + 2. ^2ah .1 + 12 – 5 ! ^2a + 1h
2 2
= =
2 2
– 5 ! ( 2 a + 1) – 5 + 2a + 1 – 5 – 2a –1 2a – 4 –2a – 6
= = , = , = (a – 2), – (a + 3)
2 2 2 2 2

1 1 2
15. Solve for x : + = , x ! 1, 2, 3. [CBSE (AI) 2016]
(x – 1) (x – 2) (x – 2) (x – 3) 3

1 1 2 (x – 3) + (x – 1) 2
Sol. + = ⇒ =
(x – 1) (x – 2) (x – 2) (x – 3) 3 (x – 1) (x – 2) (x – 3) 3
⇒ 3(x – 3 + x – 1) = 2(x – 1)(x – 2)(x – 3)
⇒ 3(2x – 4) = 2(x – 1)(x – 2)(x – 3)
⇒
3 × 2(x – 2) = 2(x – 1)(x – 2)(x – 3) ⇒ 3 = (x – 1)(x – 3) i.e., x2 – 4x = 0
⇒
x(x – 4) = 0 \ x = 0, x = 4
16. If the roots of the equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 in x are equal, then show that
either a = 0 or a3 + b3 + c3 = 3abc. [CBSE (AI) 2017]
Sol. For equal roots D = 0
Therefore 4(a2 – bc)2 – 4(c2 – ab) (b2 – ac) = 0
⇒ 4[a4 + b2c2 – 2a2bc – b2c2 + ac3 + ab3 – a2bc] = 0

⇒
a(a3 + b3 + c3 – 3abc) = 0
⇒ Either a = 0 or a3 + b3 + c3 = 3abc

17. If the roots of the quadratic equation (x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0 are equal,
then show that a = b = c. [CBSE (F) 2017]
Sol. Given (x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0
⇒
x – ax – bx + ab + x2 – bx – cx + bc + x2 – cx – ax + ac = 0
2

⇒ 3x2 – 2(a + b + c)x + ab + bc + ca = 0

Now, for equal roots D = 0
⇒
B2 – 4AC = 0
⇒ 4(a + b + c)2 – 12(ab + bc + ca) = 0

Quadratic Equations 87
 ⇒ 4a2 + 4b2 + 4c2 + 8ab + 8bc + 8ca – 12ab – 12bc – 12ca = 0

⇒ 2[2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca] = 0

⇒ 2[(a2 + b2 – 2ab) + (b2 + c2 – 2bc) + (c2 + a2 – 2ca)] = 0

⇒ [(a – b)2 + (b – c)2 + (c – a)2] = 0

⇒
a – b = 0, b – c = 0, c – a = 0
⇒
a = b, b = c, c = a ⇒ a=b=c
18. A farmer plants apple trees in a square pattern. In order to protect the apple trees against the
wind he plants conifer trees all around the orchard. Here you see a diagram of this situation
where you can see the pattern of apple trees and conifer trees for any number (n) rows of apple
trees:
= conifer
= apple tree

(a) Complete the table:

n Number of apple trees Number of conifer trees
1 1 8
2 4
3
4
5
(b) There are two formulae you can use to calculate the number of apple trees and the number
of conifer trees for the pattern described above:
Number of apple trees = n2
Number of conifer trees = 8n
Where n is the number of rows of apple trees.
There is a value of n for which the number of apple trees equals the number of conifer
trees. Find the value of n and show your method of calculating this.
(c) Suppose the farmer wants to make a much larger orchard with many rows of trees. As the
farmer makes the orchard bigger, which will increase more quickly: the number of apple
trees or the number of conifer trees? Explain how you found your answer.
Sol.: (a) Completed table as below:
n Number of apple trees Number of conifer trees
1 1 8
2 4 16
3 9 24
4 16 32
5 25 40

88 Xam idea Mathematics–X

 (b) According to question,

number of apple trees = number of conifer trees
⇒ n2 = 8n

⇒ n2 – 8n = 0
⇒ n(n – 8) = 0
⇒ n = 0 or (n – 8) = 0
⇒ n = 8, here n ≠ 0

(c) The number of apple trees increase more quickly than the number of conifer trees. We can
decide it because apple trees are increasing linearly.

Long Answer Questions [5 marks]

2 2
1. Using quadratic formula, solve the following equation for x: abx + (b – ac) x – bc = 0
Sol. We have, abx2 + (b2 – ac) x – bc = 0
Here, A = ab, B = b2 – ac, C = – bc
– B ! B2 – 4AC
\
x=
2A
– (b2 – ac) ! (b2 – ac) 2 – 4 (ab) (– bc) – (b2 – ac) ! (b2 – ac) 2 + 4ab2 c
⇒
x= ⇒ x =
2ab 2ab
– (b2 – ac) ! (b 4 – 2ab2 c + a2 c2 + 4ab2 c
⇒ x =

2ab
– (b2 – ac) ! (b2 + ac) 2 – (b2 – ac) ! (b2 + ac)
⇒ x =
⇒ x=
2ab 2ab
– (b2 – ac) + (b2 + ac) – (b2 – ac) – (b2 + ac)
⇒ x= or x =
2ab 2ab
2ac – 2b 2 c
x = or x = – b
x= or x= ⇒ b a
2ab 2ab
2. Find the value of p for which the quadratic equation
(2p + 1)x2 – (7p + 2)x + (7p – 3) = 0 has equal roots. Also find these roots. [CBSE Delhi 2014]
Sol. Since the quadratic equation has equal roots, D = 0 i.e., b2 – 4ac = 0
In (2p + 1)x2 – (7p + 2)x + (7p – 3) = 0
Here, a = (2p + 1), b = – (7p + 2), c = (7p – 3)
\ (7p + 2)2 – 4(2p + 1)(7p – 3) = 0

⇒ 49p2 + 4 + 28p – (8p + 4) (7p – 3) = 0

⇒ 49p2 + 4 + 28p – 56p2 + 24p – 28p + 12 = 0

⇒ –7p2 + 24p + 16 = 0
⇒ 7p2 – 24p – 16 = 0
⇒ 7p2 – 28p + 4p – 16 = 0
⇒ 7p(p – 4) + 4(p – 4) = 0
4
⇒ (7p + 4)(p – 4) = 0
⇒ p=– or p = 4
7
4
For p=–
7

d2 # d n + 1 n x2 – d 7 # d n + 2 n x + d7 # d n – 3n = 0
–4 –4 –4

7 7 7
–1
⇒
x2 + 2x – 7 = 0 ⇒ x2 – 14x + 49 = 0
7
⇒
x2 – 7x – 7x + 49 = 0 ⇒ x(x – 7) – 7(x – 7) = 0

Quadratic Equations 89
 ⇒ (x – 7)2 = 0
⇒ x = 7, 7
2
For p = 4, (2 × 4 + 1)x – (7 × 4 + 2)x + (7 × 4 – 3) = 0
⇒ 9x2 – 30x + 25 = 0
⇒ 9x2 – 15x – 15x + 25 = 0
⇒ 3x(3x – 5) –5(3x – 5) = 0 ⇒ (3x – 5) (3x – 5) = 0
⇒ 5 5
x= ,
3 3
1
3. The sum of the reciprocals of Rehman’s age (in years) 3 years ago and 5 years from now is .
3
Find his present age. [NCERT]
Sol. Let the present age of Rehman be x years.
So, 3 years ago, Rehman’s age = (x – 3) years
And 5 years from now, Rehman’s age = (x + 5) years
Now, according to question, we have
1 1 1
+ =
x–3 x+5 3
x+5+x–3 1 2x + 2 1
⇒
= ⇒ =
(x – 3) ( x + 5) 3 (x – 3) ( x + 5) 3
⇒ 6x + 6 = (x – 3) (x + 5) ⇒ 6x + 6 = x2 + 5x – 3x – 15
⇒ x2 + 2x – 15 – 6x – 6 = 0 ⇒ x2 – 4x – 21 = 0
⇒ x2 – 7x + 3x – 21 = 0 ⇒ x(x – 7) + 3(x – 7) = 0
⇒ (x – 7) (x + 3) = 0 ⇒ x = 7 or x = –3
But x ≠ –3 (age cannot be negative)
Therefore, present age of Rehman = 7 years.
1
4. The difference of two natural numbers is 5 and the difference of their reciprocals is . Find
10
the numbers. [CBSE Delhi 2014]
Sol. Let the two natural numbers be x and y such that x > y.
According to the question,
Difference of numbers, x – y = 5 ⇒ x = 5 + y ...(i)
Difference of the reciprocals,
1 1 1
y – x = 10 ...(ii)
Putting the value of (i) in (ii)
1 1 1 5+y– y 1
–
y 5+y = ⇒ =
10 y ( 5 + y) 10

⇒
50 = 5y + y2 ⇒ y2 + 5y – 50 = 0
2
⇒
y + 10y – 5y – 50 = 0 ⇒ y(y + 10) – 5(y + 10) = 0
⇒ (y – 5) (y + 10) = 0

\
y = 5 or y = –10

 y is a natural number. \ y=5
Putting the value of y in (i), we have
x = 5 + 5 ⇒ x = 10
The required numbers are 10 and 5.

90 Xam idea Mathematics–X

 5. The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
 [CBSE (F) 2014, Delhi 2017 (C)]
Sol. Let the two consecutive odd numbers be x and x + 2.
⇒ x2 + (x + 2)2 = 394 ⇒ x2 + x2 + 4 + 4x = 394
⇒ 2x2 + 4x + 4 = 394 ⇒ 2x2 + 4x – 390 = 0
⇒
x2 + 2x – 195 = 0 ⇒ x2 + 15x – 13x – 195 = 0
⇒
x(x + 15) – 13(x + 15) = 0 ⇒ (x – 13) (x + 15) = 0
⇒
x – 13 = 0 or x + 15 = 0 ⇒ x = 13 or x = –15
Hence, the numbers are 13 and 15 or –15 and –13.
6. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2
marks more in Mathematics and 3 marks less in English, the product of her marks would have
been 210. Find her marks in the two subjects. [NCERT]
Sol. Let Shefali’s marks in Mathematics be x.
Therefore, Shefali’s marks in English is (30 – x).
Now, according to question,
(x + 2) (30 – x – 3) = 210 ⇒ (x + 2) (27 – x) = 210
⇒ 27x – x2 + 54 – 2x = 210 ⇒ 25x – x2 + 54 – 210 = 0
⇒ 25x – x2 – 156 = 0 ⇒ – (x2 – 25x + 156) = 0
⇒ x2 – 25x + 156 = 0 ⇒ x2 – 13x – 12x + 156 = 0
⇒ x(x – 13) – 12(x – 13) = 0 ⇒ (x – 13) (x – 12) = 0
Either x – 13 = 0 or x – 12 = 0
⇒
x = 13 or x = 12
∴
Shefali’s marks in Mathematics = 13, marks in English = 30 – 13 = 17
or Shefali’s marks in Mathematics = 12, marks in English = 30 – 12 = 18.
7. A train travels 360 km at a uniform speed. If the speed has been 5 km/h more, it would have
taken 1 hour less for the same journey. Find the speed of the train. [CBSE 2019 (30/5/1)]
Sol. Let the uniform speed of the train be x km/h.
360
Then, time taken to cover 360 km = x h
Now, new increased speed = (x + 5) km/h
360
So, time taken to cover 360 km = h
x+5
360 360
According to question, x – x + 5 = 1
360 (x + 5 – x)
360 c x – m = 1
1 1
⇒
⇒ =1
x+5 x (x + 5 )
360 # 5
⇒
= 1 ⇒ 1800 = x2 + 5x
x ( x + 5)
\ x2 + 5x – 1800 = 0 ⇒ x2 + 45x – 40x – 1800 = 0
⇒
x(x + 45) – 40 (x + 45) = 0 ⇒ (x + 45) (x – 40) = 0
Either x + 45 = 0 or x – 40 = 0
\
x = – 45 or x = 40
But x cannot be negative, so x ≠ – 45
Therefore, x = 40
Hence, the uniform speed of train is 40 km/h.

Quadratic Equations 91
 8. The sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m,
find the sides of the two squares.
Sol. Let x be the length of the side of first square and y be the length of side of the second square.
Then, x2 + y2 = 468 ...(i)
Let x be the length of the side of the bigger square.
4x – 4y = 24
⇒ x – y = 6
or x = y + 6 ...(ii)
Putting the value of x in terms of y from equation (ii) in equation (i), we get
(y + 6)2 + y2 = 468
⇒
y2 + 12y + 36 + y2 = 468 or 2y2 + 12y – 432 = 0
⇒
y2 + 6y – 216 = 0 ⇒ y 2 + 18y – 12y – 216 = 0
⇒
y(y + 18) –12(y + 18) = 0 ⇒ (y + 18)(y – 12) = 0
Either y + 18 = 0 or y – 12 = 0
⇒
y = –18 or y = 12
But, sides cannot be negative, so y = 12
Therefore, x = 12 + 6 = 18
Hence, sides of two squares are 18 m and 12 m.
9. Seven years ago Varun’s age was five times the square of Swati’s age. Three years hence,
Swati’s age will be two-fifth of Varun’s age. Find their present ages.
Sol. Seven years ago, let Swati’s age be x years. Then, seven years ago Varun’s age was 5x2 years.
\
Swati’s present age = (x + 7) years
Varun’s present age = (5x2 + 7) years
Three years hence,
Swati’s age = (x + 7 + 3) years = (x + 10) years
Varun’s age (5x2 + 7 + 3) years = (5x2 + 10) years
According to the question,
2 2
x + 10 = (5x2 + 10) ⇒ x + 10 = × 5 (x2 + 2)
5 5
⇒
x + 10 = 2x2 + 4 ⇒ 2x2 – x – 6 = 0
⇒ 2x2 – 4x + 3x – 6 = 0
⇒ 2x (x – 2) + 3(x – 2) = 0
⇒ (2x + 3) (x – 2) = 0

⇒
x – 2 = 0 or 2x + 3 = 0
⇒
x = 2 [ 2x + 3 ≠ 0 as x > 0]
Hence, Swati’s present age = (2 + 7) years = 9 years
and Varun’s present age = (5 × 22 + 7) years = 27 years
10. A train covers a distance of 300 km at a uniform speed. If the speed of the train is increased
by 5 km/hour, it takes 2 hours less in the journey. Find the original speed of the train.
[CBSE (AI) 2017]
Sol. Let the original speed of the train = x km/h.
300
Therefore, time taken to cover 300 km = x hours ...(i)

When its speed is increased by 5 km/h, then time taken by the train to cover the distance of
300
300 km = hours ...(ii)
x+5

92 Xam idea Mathematics–X

 300 300
According to the question, c x – m hours = 2 hours
x+5
300 (x + 5) – 300x 300 {(x + 5) – x}
⇒ = 2 ⇒ =2
x (x + 5) x (x + 5)

⇒ 2x (x + 5) = 300 × 5
⇒ 2x2 + 10x – 1500 = 0
⇒
x2 + 5x – 750 = 0 ⇒ x2 + 30x – 25x – 750 = 0
⇒
x(x + 30) – 25(x + 30) = 0 ⇒ (x – 25) (x + 30) = 0
⇒
x = 25 or x = – 30
⇒
x = 25 ( Speed cannot be negative)
Therefore, the usual speed of the train = 25 km/h.
11. A train travels at a certain average speed for a distance of 63 km and then travels at a distance
of 72 km at an average speed of 6 km/h more than its original speed. If it takes 3 hours to
complete total journey, what is the original average speed? [CBSE 2018]
Sol. Let original average speed of the train = x km/h.
Distance travelled by train at this speed = 63 km
63
Time taken = x hours
63
Now, if the average speed is increased by 6 km/h then time taken = hours
x+6
According to question,
63 72 21 24
x + x+6 =3 & x + x + 6 = 1
⇒
21(x + 6) + 24x = x2 + 6x
⇒
x2 – 39x – 126 = 0
⇒
(x – 42)(x + 3) = 0
⇒
x = – 3 or x = 42
But x cannot be equal to –3 as speed cannot be negative.
Hence, original average speed of the train is 42 km/h.
12. A two digit number is such that the product of its digits is 18. When 63 is subtracted from the
number, the digits interchange their places. Find the number.
Sol. Let the digit at tens place be x.
18
Then, digit at unit place = x
18
\
Number = 10x + x
18
and number obtained by interchanging the digits = 10 × x + x
According to question,

a10x + x k – 63 = 10 # x + x ⇒ a10x + x k – a10 # x + x k = 63

18 18 18 18

18 180 162
⇒
10x + x – x – x = 63 ⇒ 9x – x – 63 = 0
⇒ 9x2 – 63x – 162 = 0
⇒ x2 – 7x – 18 = 0
2
⇒
x – 9x + 2x – 18 = 0 ⇒ x(x – 9) + 2(x – 9) = 0
⇒ (x – 9) (x + 2) = 0
⇒ x = 9 or x=–2
⇒
x = 9 [ a digit can never be negative]
18
Hence, the required number = 10 × 9 + = 92.
9

Quadratic Equations 93
 13. If twice the area of a smaller square is subtracted from the area of a larger square; the result is
14 cm2. However, if twice the area of the larger square is added to three times the area of the
smaller square, the result is 203 cm2. Determine the sides of the two squares.
Sol. Let the sides of the larger and smaller squares be x and y respectively. Then
x2 – 2y2 = 14 ...(i) and 2x2 + 3y2 = 203 ...(ii)
Operating (ii) –2 × (i), we get
2x2 + 3y2 – (2x2 – 4y2) = 203 – 2 × 14
⇒ 2x2 + 3y2 – 2x2 + 4y2 = 203 – 28

⇒ 7y2 = 175
⇒ y2 = 25 ⇒ y = ±5
⇒ y = 5 [ Side cannot be negative]
By putting the value of y in equation (i), we get
x2 – 2 × 52 = 14 ⇒ x2 – 50 = 14 or x2 = 64
\
x = ± 8 or x = 8
\ Sides of the two squares are 8 cm and 5 cm.

14. If Zeba was younger by 5 years than what she really is, then the square of her age (in years) would
have been 11 more than five times her actual age. What is her age now?  [NCERT Exemplar]
Sol. Let the present age of Zeba be x years.
Age before 5 years = (x – 5) years
According to given condition,
⇒ (x – 5)2 = 5x + 11
⇒ x 2 + 25 – 10x = 5x + 11
2
⇒
x – 10x – 5x + 25 – 11 = 0 ⇒ x2 – 15x + 14 = 0
⇒
x2 – 14x – x + 14 = 0 ⇒ x(x – 14) –1(x – 14) = 0
⇒ (x – 1)(x – 14) = 0
⇒ x–1=0 or x – 14 = 0
x=1 or x = 14
But present age cannot be 1 year.
\  Present age of Zeba is 14 years.
15. The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.
 [CBSE (F) 2014]
Sol. Let the two consecutive multiples of 7 be x and x +7.
x2 + (x + 7)2 = 637
⇒
x2 + x2 + 49 + 14x = 637 ⇒ 2x2 + 14x – 588 = 0
⇒
x2 + 7x – 294 = 0 ⇒ x2 + 21x – 14x – 294 = 0
⇒
x(x + 21) – 14(x + 21) = 0 ⇒ (x – 14) (x + 21) = 0
x = 14 or x = –21
\  The multiples are 14 and 21 or –21 and –14.

1 2 4
16. Solve for x: + = , x ! –1, – 2, – 4 [CBSE (AI) 2016]
+
x 1 +
x 2 +
x 4
1 2 4
Sol. + =
+
x 1 +
x 2 x+4
x + 2 + 2 ( x + 1) 4
⇒
= ⇒ (x + 4) (x + 2 + 2x + 2) = 4(x + 1)(x + 2)
( x + 1) ( x + 2) x+4
⇒ (x + 4) (3x + 4) = 4(x2 + 3x + 2) ⇒ x2 – 4x – 8 = 0

4 ! 16 + 32 4!4 3
⇒
x= = = 2!2 3
2 2

94 Xam idea Mathematics–X

 17. Find the positive value(s) of k for which both quadratic equations x2 + kx + 64 = 0 and
x2 – 8x + k = 0 will have real roots. [CBSE (F) 2016]
Sol. (i) For x2 + kx + 64 = 0 to have real roots
k2 – 4(1) (64) ≥ 0
i.e., k2 – 256 ≥ 0 ⇒ (k – 16) (k + 16) ≥ 0   ⇒   k ≤ –16 or k ≥ 16
2
(ii) For x – 8x + k = 0 to have real roots
(–8)2 – 4(k) ≥ 0 i.e., 64 – 4k ≥ 0 ⇒ k ≤ ± 16
For (i) and (ii) to hold simultaneously k = 16.
18. Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same
point in 4 hours 30 minutes. Find the speed of the stream. [CBSE Delhi 2017]
Sol. Let the speed of stream be x km/h.
∴  Speed of boat in upstream = (15 – x) km/h

   Speed of boat in downstream = (15 + x) km/h
According to question,
30 30 1 9
+ =4 =
15 – x 15 + x 2 2
30 (15 + x + 15 – x) 9
⇒
=
(15 – x) (15 + x) 2
⇒
30 × 2 × 30 = 9(225 – x2) ⇒   100 × 2 = 225 – x2
⇒
200 = 225 – x2
⇒
x2 = 25 ⇒   x = ±5
⇒
x = 5 (Rejecting – 5)
∴
Speed of stream = 5 km/h
19. A motorboat whose speed in still water is 9 km/h, goes 15km downstream and comes back to the
same spot, in a total time of 3 hours 45 minutes. Find the speed of the stream.[CBSE 2019 (30/3/3)]
Sol. Let speed of the stream be x km/h.
Speed of boat in upstream = (9 – x) km/h
Speed of boat in downstream = (9 + x) km/h
According to question,
15 15 3
+ =3
+
9–x 9 x 4
15 15 15
⇒
+ =
9 – x 9+ x 4
15 [9 + x + 9 – x] 15
⇒
=
(9 – x) (9 + x) 4

⇒
15 × 4 × 18 = 15(81 – x2)
⇒
72 = 81 – x2
⇒
x2 = 9
⇒
x =  3, but x  –3 (speed can’t be negative)
 x = 3 ⇒ speed of stream = 3 km/h
20. Ram takes 6 days less than Bhagat to finish a piece of work. If both of them together can finish
the work in 4 days, in how many days Bhagat alone can finish the work. [CBSE Delhi 2017(C)]
Sol. Let Bhagat alone can do the work in x number of days.
∴ Ram takes (x – 6) number of days.

Quadratic Equations 95
 1
Work done by Bhagat in 1 day =
x
1
Work done by Ram in 1 day =
x–6
According to the question,
1 1 1
+ =
x x–6 4
x–6+x 1
⇒ = ⇒ 4(2x – 6) = x(x – 6)
x ( x – 6) 4
⇒
x2 – 14x + 24 = 0 ⇒ x2 – 12x – 2x + 24 = 0
⇒
x (x – 12) – 2 (x – 12) = 0
⇒
(x – 12) (x – 2) = 0 ⇒ Either x – 12 = 0 or x – 2 = 0
⇒
x = 12, as x ≠ 2

 Bhagat alone can do the work in 12 days.

HOTS [Higher Order Thinking Skills]

1. One-fourth of a herd of camels was seen in the forest. Twice the square root of the herd had
gone to mountains and the remaining 15 camels were seen on the bank of a river. Find the total
number of camels.
Sol. Let x be the total number of camels.
x
Then, number of camels in the forest =
4
Number of camels on mountains = 2 x
and number of camels on the bank of river = 15
x
Thus, total number of camels = + 2 x + 15
4
Now by hypothesis, we have
x
+ 2 x + 15 = x ⇒ 3x – 8 x – 60 = 0
4

Let x = y , then x = y2
⇒ 3y2 – 8y – 60 = 0
⇒ 3y2 – 18y + 10y – 60 = 0
⇒ 3y(y – 6) + 10(y – 6) = 0 ⇒ (3y + 10)(y – 6) = 0

10
⇒
y=6 or y=–
3
10 10 2 100
Now, y =–
3 ⇒ x = c– m = ( x = y2)
3 9
But, the number of camels cannot be a fraction.
\
y = 6 ⇒ x = 62 = 36
Hence, the number of camels = 36
2. Solve the following quadratic equation:
9x2 – 9 (a + b) x + [2a2 + 5ab + 2b2] = 0 [CBSE (F) 2016]
2 2 2
Sol. Consider the equation 9x – 9 (a + b) x + [2a + 5ab + 2b ] = 0
2
Now comparing with Ax + Bx + C = 0, we get
A = 9, B = – 9 (a + b) and C = [2a2 + 5ab + 2b2]
Now discriminant,

96 Xam idea Mathematics–X

 D = B2 – 4AC
= {– 9 (a + b)}2 – 4 × 9 (2a2 + 5ab + 2b2) = 92 (a + b)2 – 4 × 9 (2a2 + 5ab + 2b2)
= 9 {9 (a + b)2 – 4 (2a2 + 5ab + 2b2)} = 9 {9a2 + 9b2 + 18ab – 8a2 – 20ab – 8b2}
= 9 {a2 + b2 – 2ab} = 9 (a – b)2
Now using the quadratic formula,

– B! D 9 ( a + b) ! 9 ( a – b ) 2
x= , we get x =
2A 2#9
9 ( a + b) ! 3 ( a – b) 3 ( a + b) ! ( a – b)
⇒
x= ⇒ x=
2#9 6
( 3 a + 3 b) + ( a – b) ( 3 a + 3 b) – ( a – b)
⇒
x= and x=
6 6
( 4 a + 2 b) ( 2 a + 4 b)
⇒ x= and x=
6 6
2a + b a + 2b
⇒
x= and x= are required solutions.
3 3

1
3. Two taps running together can fill a tank in 3 hours. If one tap takes 3 hours more than the
13
other to fill the tank, then how much time will each tap take to fill the tank? [CBSE (AI) 2017]
Sol. Let, time taken by faster tap to fill the tank be x hours.
Therefore, time taken by slower tap to fill the tank = (x + 3) hours
Since the faster tap takes x hours to fill the tank.
1
\ Portion of the tank filled by the faster tap in one hour
=
x
1
Portion of the tank filled by the slower tap in one hour =
x+3

1 13
Portion of the tank filled by the two tap together in one hour = =
40 40
13
According to question,
1 1 13 x+3+x 13
⇒
x + x + 3 = 40 ⇒
+
x ( x 3)
=
40
⇒
40 (2x + 3) = 13x (x + 3) ⇒ 80x + 120 = 13x2 + 39x
⇒
13x2 – 41x – 120 = 0 ⇒ 13x2 – 65x + 24x – 120 = 0
⇒
13x (x – 5) + 24 (x – 5) = 0 ⇒ (x – 5) (13x + 24) = 0
Either x – 5 = 0 or 13x + 24 = 0
– 24
⇒ x=5 or x=
13
⇒
x = 5 [ x cannot be negative]
Hence, time taken by faster tap to fill the tank = x = 5 hours
and time taken by slower tap = x + 3 = 5 + 3 = 8 hours.

Quadratic Equations 97
 4. In a rectangular park of dimensions 50 m × 40 m, a rectangular pond is constructed so that the
area of grass strip of uniform width surrounding the pond would be 1184 m2. Find the length
and breadth of the pond. [NCERT Exemplar, CBSE (F) 2017]

Sol. Let ABCD be rectangular lawn and EFGH be rectangular pond. Let x m be the width of grass
area, which is same around the pond.
Given, Length of lawn = 50 m
Width of lawn = 40 m
⇒
Length of pond = (50 – 2x) m
Breadth of pond = (40 – 2x) m
Also given,
Area of grass surrounding the pond = 1184m2
Fig. 4.1
⇒
Area of rectangular lawn – area of pond = 1184 m2
⇒
50 × 40 – {(50 – 2x) × (40 – 2x)} = 1184
⇒
2000 – (2000 – 80x – 100x + 4x2) = 1184
⇒
2000 – 2000 + 180x – 4x2 = 1184
⇒
4x2 – 180x + 1184 = 0 ⇒ x2 – 45x + 296 = 0
⇒
x2 – 37x – 8x + 296 = 0 ⇒ x(x – 37) – 8(x – 37) = 0
⇒ (x – 37) (x – 8) = 0
⇒ x – 37 = 0 or x – 8 = 0
⇒
x = 37 or x = 8

x = 37 is not possible as in this case length of pond becomes 50 – 2 × 37 = – 24 (not possible)
Hence, x = 8 is acceptable
\
Length of pond = 50 – 2 × 8 = 34 m
Breadth of pond = 40 – 2 × 8 = 24 m
5. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side
is 30 metres more than the shorter side, find the sides of the field. [NCERT]
Sol. Let ABCD be the rectangular field. Let the shorter side BC of the rectangle = x metres.
According to question,
Diagonal of the rectangle, AC = (x + 60) metres
side of the rectangle, AB =(x + 30) metres
By Pythagoras theorem, AC2 = AB2 + BC2
\
(x + 60)2 = (x + 30)2 + x2
or x2 + 120x + 3600 = x2 + 60x + 900 +x2
\ (2x2 – x2) + (60x – 120x) + 900 – 3600 = 0

Fig. 4.2
or x2 – 60x – 2700 = 0
or (x – 90)(x + 30) = 0
\ Either x – 90 = 0 or x + 30 = 0

⇒
x = 90 or x = – 30 (But side cannot be negative)
So, the shorter side of rectangle = 30 m
and longer side of rectangle = 130 m

98 Xam idea Mathematics–X

 PROFICIENCY EXERCISE
QQ Objective Type Questions: [1 mark each]
1. Choose and write the correct option in each of the following questions.

(i) The discriminant of the equation bx2 + ax + c = 0, b  0 is given by

(a) b2 – 4ac (b) a2 + 4bc (c) a2 – 4bc (d) b2 + 4ac
16
(ii) The roots of the equation x + x = 10 are
(a) 4, 6 (b) 4, 4 (c) 4, 5 (d) 2, 8
(iii) If the roots of ax2 + bx + c = 0 are equal, then the value of c is
–b b –b2 b2
(a) (b) (c) (d)
2a 2a 4a 4a
2
(iv) The value of k for which x = –2 is a root of the equation kx + x – 6 = 0 is
–3
(a) (b) –1 (c) –2 (d) 2
2
(v) If the discriminant of a quadratic equation is less than zero then it has
(a) equal roots (b) real roots (c) no real roots (d) can't be determined
2. Fill in the blanks.
2
(i) The value of k is _____________ if one root of the quadratic equation 6x2 – x – k = 0 is .
3
(ii) A quadratic equation has _____________ roots if its discriminant is greater than zero.
(iii) Every quadratic equation has _____________ two roots.
(iv) The quadratic equation 3x2 – 7x + 4 = 0 has _____________ roots.
(v) If roots of a quadratic equation are equal, then its discriminant is equal to _____________.
QQ Very Short Answer Questions: [1 mark each]
2
3.
If one root of 5x + 13x + k = 0 is the reciprocal of the other root, then find value of k.
 [CBSE 2018 (C)]
2
4.
For what values of k, the roots of the equation x + 4x + k = 0 are real? [CBSE 2019 (30/1/1)]
2
5.
Find the value of k for which the roots of the equation 3x – 10x + k = 0 are reciprocal of each
other. [CBSE 2019 (30/1/1)]

Find the nature of roots of the quadratic equation 2x2 – 4x + 3 = 0.

6. [CBSE 2019 (30/2/1)]
2
7.
Find the nature of the roots of the quadratic equation 4x + 4 3 x + 3 = 0. [CBSE 2019 (30/2/3)]
For what values of k does the quadratic equation 4x2 – 12x – k = 0 have no real roots?
8.
[CBSE 2019 (30/4/2)]
2
9.
Find the value(s) of k for which the quadratic equation 3x + kx + 3 = 0 has real and equal roots.
 [CBSE 2019 (30/5/1)]
2
10. For what values of 'a' the quadratic equation 9x – 3ax + 1 = 0 has equal roots?
 [CBSE 2019 (C) (30/1/1)]
2 1
11. If one root of the quadratic equation 2x + 2x + k = 0 is – , then find the value of k.
3
[CBSE 2019 (C) (30/1/1)]
12. What is the value of k for which 3 is a root of the equation kx2 – 7x + 3 = 0?
13. If the equation 16x2 + 6kx + 4 = 0 has equal roots, then what is the value of k?

Quadratic Equations 99
 14. If ax2 + bx + c = 0 has equal roots, then what is b equal to?
15. What can be said about the quadratic equation x2 – 5x + 3 = 0?
16. What are the roots of the equation x2 – 9x + 20 = 0?
17. What is the value(s) of k for which the equation kx2 – kx + 1 = 0 has equal roots?
18. If the roots of the equation (a2 + b2) x2 – 2b(a + c)x + (b2 + c2) = 0 are equal, then what is the
value of b2?
19. What is the discriminant of the quadratic equation 3 3 x2 + 10x + 3 = 0 ?
20. Write a quadratic equation which has the product of two roots as 5.
2
21. If one root of the quadratic equation 6x2 – x – k = 0 is , then find the value of k.
3
[CBSE (F) 2017]
QQ Short Answer Questions-I: [2 marks each]
State whether the following quadratic equations have two distinct real roots or not. Justify your answer.
(Q. 22 – 23)
22. 2x2 + x – 1 = 0
23. (x + 4)2 – 8x = 0 [NCERT Exemplar]
24. Write the condition to be satisfied for which equations ax + 2bx + c = 0 and bx2 – 2 ac x + b = 0
2

have equal roots.

25. If equation ax2 + bx + c = 0 has equal roots, then find ‘c’ in terms of ‘a’ and ‘b’.
QQ Short Answer Questions-II: [3 marks each]
2
26. Write all the values of p for which the quadratic equation x + px + 16 = 0 has equal roots. Find
the roots of the equation so obtained. [CBSE 2019, (30/2/1)]
27. Sum of the areas of two squares is 157 m2. If the sum of their perimeters is 68 m, find the sides
of the two squares. [CBSE 2019, (30/4/2)]
28. A motorboat whose speed is 18 km/h in still water takes one hour more to go 24 km upstream
than to return downstream to the same spot. Find the speed of the stream.
[CBSE 2018, 2019 (30/4/3), 2020 (30/5/1)]
29. Find the value of k for which the quadratic equation (k +1)x2 – 6(k + 1)x + 3(k + 9) = 0, k  –1
has equal roots. [CBSE 2019 (C) (30/1/3)]
Find the roots of the following quadratic equations using the quadratic formula (Q. 30 – 31)
30. –3x2 + 5x + 12 = 0
31. x2 – 3 5 x + 10 = 0
14 5
32. Solve for x: −1 = ; ≠ −3, −1 [CBSE Delhi 2014]
x+3 x +1
3 1 2 1
33. Solve for x: – = ; x ! 1, x ! [CBSE Delhi 2014]
+
x 1 2 3x – 1 3
Find the roots of the following quadratic equations by the factorisation method (Q. 34 – 37)

34. 3x2 + 5 5 x – 10 = 0
5
35. 2x2 + x – 2 = 0 [NCERT Exemplar]
3
36. 4x2 – 4a2x + (a4 – b4) = 0 [CBSE Delhi 2015]

100 Xam idea Mathematics–X

 x –1 2x + 1 1
37. + = 2, x ! – , 1 [CBSE (AI) 2017]
2x + 1 x –1 2
38. Find two consecutive positive integers, sum of whose squares is 925.
5
39. The sum of two numbers is 20 and the sum of their reciprocals is . Find the numbers.
24
40. The difference of squares of two numbers is 88. If the larger number is 5 less than twice the
smaller number, then find the two numbers.
41. If one root of the quadratic equation 2x2 + kx – 6 = 0 is 2, find the value of k. Also find the other root.

42. If the roots of the equation (a2 + b2)x2 – 2(ac + bd) x + (c2 + d2) = 0 are equal, prove that ad = bc.
[CBSE 2019 (C), (30/1/1)]
OR
Show that if the roots of the following quadratic equation are equal, then ad = bc
x2(a2 + b2) + 2(ac + bd) x + (c2 + d2) = 0 [CBSE (AI) 2017, Delhi 2017 (C)]
43. Seven years ago, Varun’s age was five times the square of Swati’s age. Three years hence, Swati’s
age will be two fifth of Varun’s age. Find their present ages.
44. The area of a right angled triangle is 165 m2. Determine its base and altitude if the later exceeds
the former by 7 m.
45. Is it possible to design a rectangular park of perimeter 40 m and area 400 m2? If so, find its
length and breadth.
46. Determine the condition for the roots of the equation ax2 + bx + c = 0 to differ by 2.
47. Show that the equation 2(a2 + b2) x2 + 2(a + b)x + 1 = 0 has no real roots, when a ≠ b.
48. Find the roots of the equation: ax2 + a = a2x + x
49. If 2 is a root of the quadratic equation 3x2 + px – 8 = 0 and the quadratic equation 4x2 – 2px + k = 0
has equal roots, find the value of k. [CBSE (F) 2014]
50. Find that non-zero value of k, for which the quadratic equation kx2 + 1 – 2(k – 1)x + x2 = 0 has
equal roots. Hence find the roots of the equation. [CBSE Delhi 2015]
QQ Long Answer Questions: [5 marks each]
51. Solve for x:
1 1 1 1 –2a – b
= + + ; x ! 0, x ! , a, b ! 0 [CBSE 2019 (30/4/3)]
2a + b + 2x 2a b 2x 2

52. The sum of the areas of two squares is 640 m2. If the difference of their perimeters is 64 m, find
the sides of the squares. [CBSE 2019 (30/4/3)]
53. Solve for x:
1 1 1 1
= a + + x ; a ! b ! 0 , x ! 0 , x ! – ( a + b)  [CBSE 2019 (30/5/1)]
a+b+x b

54. Solve the following equation for x: [CBSE 2019(C)(30/1/1)]

1 2 7
+ = , x ! –1, –2, –5
+
x 1 +
x 2 x+5
55. A motorboat whose speed is 18 km/h in still water takes 1 h 30 minutes more to go 36 km
upstream than to return downstream to the same spot. Find the speed of the stream.
 [CBSE 2019(C)(30/1/3)]

Quadratic Equations 101

Find whether the following equations have real roots. If yes find them. (Q. 56 to 60)
56. 5x2 – 6x – 2 = 0
57. 8x2 + 2x – 3 = 0
1 1 3
58. + = 1, x ! , 5 [NCERT Exemplar]
2x – 3 x–5 2

59. 3x2 – 2x + 2 = 0
1 1
60. 4x2 – x+ =0
5 32
x–3 x –5 10
61. Solve for x: + = ; x ! 4, 6 [CBSE (AI) 2014]
x–4 x–6 3
x–2 x–4 10
62. Solve for x: + = ; x ≠ 3, 5 [CBSE (AI) 2014]
x–3 x–5 3
63. The sum of the squares of two consecutive even numbers is 340. Find the numbers. [CBSE (F) 2014]
2 3 23
64. Solve for x: + = , x ≠ 0, –1, 2 [CBSE Delhi 2015]
x + 1 2 ^ x – 2h 5x
5
65. The difference of two natural numbers is 5 and the difference of their reciprocals is . Find the
14
numbers. [CBSE Delhi 2014]
66. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given
number. [NCERT Exemplar]
67. At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When
Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the
present age of Nisha. Find the present ages of both Asha and Nisha. [NCERT Exemplar]
68.
There is a square field whose side is 44 m. A square flower bed is prepared in its centre leaving
a gravel path all round the flower bed. The total cost of laying the flower bed and gravelling the
path at ™2.75 and ™1.50 per m2 respectively, is ™4904. Find the width of gravel path.

69. At t minutes past 2 pm, the time needed by the minutes hand of a clock to show 3 pm was found
t2
to be 3 minutes less than minutes. Find t. [NCERT Exemplar]
4
70. A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/h more, it
would have taken 30 minutes less for the journey. Find the original speed of the train.
71. In a class test, the sum of the marks obtained by Puneet in Mathematics and Science is 28. Had
he got 3 marks more in Mathematics and 4 marks less in Science, the product of their marks,
would have been 180. Find his marks in two subjects.
72. If the list price of a toy is reduced by ™2, a person can buy 2 toys more for ™360. Find the original
price of the toy.
3 4 29 1
73. Solve for x: + = ; x ! 1, – 1, [CBSE Delhi 2015]
x+1 x –1 4x – 1 4

74. Solve for x: 4x2 + 4bx – (a2 – b2) = 0 [CBSE (F) 2017]
a b 2c
75. Find x in terms of a, b and c: + = , x ≠ a, b, c. [CBSE Delhi 2016]
x–a x–b x–c
76. Solve for x:
1 3 5 1
+ = , x ! –1, – , –4 [CBSE (AI) 2017]
+ +
x 1 5x 1 +
x 4 5

102 Xam idea Mathematics–X

 77. Solve for x:
x+3 1– x 17
– = ; x ! 0, 2 [CBSE (AI) 2017]
x–2 x 4
78. A faster train takes one hour less than a slower train for a journey of 200 km. If the speed of
slower train is 10 km/h less than that of faster train, find the speeds of two trains.
 [CBSE 2018(C)]
7
79. Two water taps together can fill a tank in 1 hours. The tap with longer diameter takes 2 hours
8
less than the tap with smaller one to fill the tank separately. Find the time in which each tap can
fill the tank separately. [CBSE 2019(30/1/1)]
80. The total cost of a certain length of a piece of cloth is `200. If the piece was 5 m longer and each
metre of cloth costs `2 less, the cost of the piece would have remained unchanged. How long is
the piece and what is its original rate per metre? [CBSE 2019(30/2/2)]
81. A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km
away on time, it has to increase its speed by 250 km/h from its usual speed. Find the usual speed
of the plane. [CBSE 2019(30/4/2)]
82. Find the dimensions of a rectangular park whose perimeter is 60 m and area is 200 m2.
 [CBSE 2019(30/4/2)]

Answers
1. (i) (c) (ii) (d) (iii) (d) (iv) (d) (v) (c)

2. (i) 2 (ii) real (iii) atmost (iv) real (v) zero

3. k = 5 4. k ≤ 4 5. k = 3 6. No real root 7. Real and equal 8. k < –9

4 8
9. k =  6 10. a =  2 11. k = 12. 2 13. ± 14. ±2 ac
9 3
15. It has two distinct real roots 16. 4 and 5 17. k = 0, 4 18. ac 19. 64

20. x2 – 2 5x + 5 = 0 21. k = 2 22. Yes, D > 0 23. No, D < 0 24. b2 = ac

b2
25. c = 26. p = – 4, 4 27. Length of side of squares are 6 m and 11 m
4a
–4
28. Speed of stream 6 km/h 29. k = 3, 30. ,3 31. 5 , 2 5 32. x = 1 and 4
3
5 –3 2
33. x = 1 and 3 34. x = , −2 5 35. x = ,
3 2 3
a2 + b2 a2 – b2
36. x = , 37. –2, –2 38. 22 and 21 39. 8 and 12 40. 9 and 13
2 2
–3
41. k = –1, other root =
2
43. Swati’s age = 9 years; Varun’s age = 27 years 44. Base=15 m, Altitude=22 cm 45. No
1 1 1
46. b2 – 4ac = 4a2 48. a, a 49. k = 1 50. k = 3, x = ,
2 2
b
51. x = – a or – 52. Length of sides of squares are 24 m and 8 m
2
–3
53. x = – a, x = – b 54. x = 1, x = 55. Speed of stream = 6 km/h
2
3 ± 19 −3 1 8 ±3 2
56. Yes; 57. Yes; , 58. Yes;
5 4 2 2

Quadratic Equations 103

 9 7
59. No 60. No real roots 61. and 7 62. and 6 63. 12 and 14
2 2
−23
64. 4 and 65. 7 and 2 66. 12
11
67. Nisha’s age is 5 years, Asha’s age is 27 years 68. 2 m 69. 14
70. 45 km/h 71. Mathematics: 9 and Science: 19 or Mathematics: 12 and Science: 16
a–b –a–b 2ab – ac – bc –11
72. ™20 73. x = –7, 4 74. , 75. x = ,0 76. ,1
2 2 a + b – 2c 17
–2
77. 4, 78. Speed of faster train = 50 km/h and speed of slower train = 40 km/h
9
79. 5 hours, 3 hours 80. l = 20 m, `10 per metre
81. Speed of plane = 750 km/h 82. Length = 20 m, breath = 10 m

SELF-ASSESSMENT TEST
Time allowed: 1 hour Max. marks: 40

Section A
1. Choose and write the correct option in the following questions. (4 × 1 = 4)
2
(i) The smallest positive value of k for which the equation x + kx + 9 = 0 has real roots is

(a) – 6 (b) 6 (c) – 36 (d) – 3
(ii) If ax2 + bx + c = 0 has equal roots, then a is equal to
b2 – b2 b –b
(a) (b) (c) (d)
4c 4c 2c 2c
(iii) The value of k for which x = – 2 is a root of the quadratic equation kx2 + x – 6 = 0 is
–3
(a) – 1 (b) – 2 (c) 2 (d)
2
(iv) If px2 + 3x + q = 0 has two roots x = – 1 and x = – 2, the value of q – p is
(a) – 1 (b) 1 (c) 2 (d) – 2
2. Fill in the blanks.  (3 × 1 = 3)
2
(i) The degree of polynomial 2x + bx + c is _______________.
(ii) The equation 16x2 + 6kx + 4 = 0 has equal roots, then the value of k is _______________ .
(iii) The roots of the equation x2 – 9x + 20 = 0 are _______________ .
3. Solve the following questions. (3 × 1 = 3)
2 2 2 2 2
(i) What is the condition that the equation (a + b ) x + 2 (ac + bd) x + c + d = 0 has no real
roots?
(ii) What are the roots of the quadratic equation 6x2 – x – k = 0 if value of k is equal to 2?
4 1
(iii) Check whether 2 + x – 5 = 0 is a quadratic equation or not? If it is quadratic then write
x
in which variable it is quadratic?

Section B
QQ Solve the following questions. (3 × 2 = 6)
x+3 3x – 7 3
4. Solve for x: = , x ! –2,  [CBSE Delhi 2017 (C)]
x+2 2x – 3 2
5. Find the value of k such that the equation (k – 12) x2 + 2(k – 12) x + 2 = 0 has equal roots.
 [CBSE Delhi 2017 (C)]
6. Solve for x: 3 x2 + 10x + 7 3 = 0  [CBSE (F) 2017]

104 Xam idea Mathematics–X

QQ Solve the following questions. (3 × 3 = 9)
3
7. The difference of two natural numbers is 3 and the difference of their reciprocals is . Find
28
the numbers. [CBSE Delhi 2014]
8. Find the value of c for which the quadratic equation 4x2 – 2(c + 1)x + (c + 1) = 0 has equal roots,
which are real. [CBSE Delhi 2017 (C)]

1 1 1 1
9. Solve for x: = a + + x ; a + b + x ≠ 0, a, b, x ≠ 0 [CBSE (F) 2017, 2018 (C)]
a+b+x b
QQ Solve the following questions. (3 × 5 = 15)
10. Three hundred apples are distributed equally among a certain number of students. Had there
been 10 more students, each would have received one apple less. Find the number of students.
11. Two water taps together can fill a tank in 9 hours 36 minutes. The tap of larger diameter takes
8 hours less than the smaller one to fill the tank separately. Find the time in which each tap can
separately fill the tank. [CBSE (F) 2016]
12. A motorboat whose speed is 24 km/h in still water takes 1 hour more to go 32 km upstream than
to return downstream to the same spot. Find the speed of the stream. [CBSE (AI) 2016]

Answers
1. (i) (b) (ii) (a) (iii) (c) (iv) (b)
8
2. (i) 2 (ii) ! (iii) 4 and 5
3
2 –1 1
3. (i) ad < bc (ii) and (iii) Yes, x
3 2
–7
4. 5, –1 5. k = 14 6. ,– 3 7. 7 and 4 8. c = –1, 3
3
9. x = –a, –b
10. 50 students 11. 24 hours, 16 hours
12. 8 km/h
zzz

Quadratic Equations 105