CHAPTER NUMBER - 4

QUADRATIC EQUATIONS

KEY CONCEPTS:
 There are three methods of solving a quadratic equation.
1) Middle term Factorization
2) Completing the square method
3) By formula method
 Standard form of a quadratic equation is ax2+bx+c = 0
If ax2+bx+c=0 is a quadratic equation, then the expression, b2- 4ac is
known as the discriminant and is denoted by D or D = b2- 4ac
 Nature of roots:
1. If D = 0, the given equation has two real and equal roots .
2. If D > 0,the given equation has two real and distinct roots.
3. If D < 0, the given equation has no real roots.

SECTION - A
MULTIPLE CHOICE QUESTIONS
1. Which of the following is not quadratic equation?
a) x2 -2x +2(3-x) b) x(x+1) + 1 = (x-2) (x-5)
c) (2x-1) (x-3) = (x+5) 9x-1) d) x3- 4x2-x+1= (x-2)3
2. If 1/2 is the root of quadratic equation x2 +kx – 5/4 =0, then the value of k
is:
a) 2 b) -2 c) 1/4 d) 1/2
2
3. The roots of the quadratic equation x -3x – 10 = 0 are:
a) 2, -5 b) -2,5 c) 2 , 5 d) 6
4. Which of the following quadratic equations has the sum of its roots as 3?
a) x2 - 3x + 6=0 b)- x2-3x-3=0 c) x2 -3/ x + 1 d) 3x2 -3x + 3
5. The sum and the product of the roots of the equation 2x2 + 5x -3=0 are
respectively:
a) 5/2, -3/2 b) -5/2, 3/2 c) 5/2, 3/2 d) -5/2, -3/2
FILL IN THE BLANKS
6. If b2- 4ac =0, then roots of the given quadratic equation ---------
7. If 1 is a root of the equation ay2 +ay+3=0 and y2 +y+b=0, then the value of
ab = ------------
8. The values of k for which 2x2+kx+3 =0 has two equal and real roots is……
___________
9. If b2- 4ac < 0, the given equation has ------------ root/roots .
10. If b2- 4ac > 0, the given equation has ------------ root/roots .
ANSWER THE FOLLOWING
11. For what value of k, x=a is a solution of the equation x2-(a+b)x+k=0.
12. If -4 is a root of x2+px-4=0 and equation x2+px+q=0 has equal roots, then
find the value of p and q.
13. Find the value of p for which the equation x2+5px+16 = 0 has no real
roots.
14. Write an equation whose sum of the roots is 3 and product of roots is 2
15. Find the roots of the quadratic equation 2x(x-7) = 1
SECTION-B
16. What can you say about the nature of the roots of the equation 2x2-4x +3=0.
17. Find the value of( m) for which the equation x2+5mx+16 =0 has equal
roots.
18. Solve the quadratic equation √3x2+11x+6√3=0 by factorization method,
19. Check whether both x=2/3 and x=-1/3 are roots of 9x2-3x-2=0 or not.
20. Find the whole number which when decreased by 20 is equal to 69 times
the reciprocal of the number.
SECTION-C
21. The sum of the areas of two squares is 640m2.If the difference of their
perimeter is 64. Then find the sides of the two squares.
22. Solve the equation x2 – (√3 + 1)x + √3x = 0 by using quadratic formula.
23. Solve the following equation for x; 9x2 - 9(a+b)x + (2a2+5ab+2b2)=0.
24. Find a natural number whose squares diminished by 84 is equal to thrice
of 8 more the given number.
25. The difference of two numbers is 5 and the difference of their reciprocals
is . Find the numbers.
SECTION-D
26. A piece of cloth costs Rs 200.1f the piece were 5m longer and each m of
cloth costed Rs 2 less, the cost of piece would have remained unchanged.
What is the original length of the piece and its original rate per meter.
27. Two pipes running together can fill a tank in 11 minutes. lf one pipe
takes 5 minutes more than the other to fill the tank, find the time in
which each pipe alone would fill the tank.
28. The denominator of a Fraction is one more than twice the numerator. If
the sum of the fraction and its reciprocal is 29/10. Find the fraction.
29. At t minutes past 2pm, the time needed by the minute hand of a clock to
show 3pm was found to be 3 minutes less than t2/ 4 minutes. Find it.
30. Solve the equation: = + + ; a + b ≠ 0 by using
factorization method.
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 CHAPTER NUMBER - 4
QUADRATIC EQUATIONS
ANSWER KEY

SECTION – A
MULTIPLE CHOICE QUESTIONS
1. b) x(x + 1) +1= (x – 2)(x – 5)
2. a) 2
3. b) - 2, 5
4. a) x² – 3x + 6 = 0
5. d) - , -
FILL IN THE BLANKS

6. -

7. ab = 3
8. K= 2
9. No real roots
10. Real & distinct roots
ANSWER THE FOLLOWING
11. a 2 – (a+b) a + k = 0
⟹ a 2 – a 2 – ab + k=0

12. p (-4) = (-4) 2 - 4p – 4 = 0

⟹ 4p = 12
⟹ P=3
Also x 2 + 3x + q = 0
Now, b 2 -4ac ⟹ (3) 2 – 4q = 0
⟹q=

13. b 2 – 4ac < 0

⟹ 25p 2 – 4(16) <0
⟹ 25p 2 < 64
⟹ P2 <

<p<
14. x2 -
Equation is x 2 - 3x + 2
15. 2x (x – 7) = 1
⟹ 2x 2 -14x-1=0

⟹x=

⟹ x=

16. b 2 – 4ac
= (-4) 2 – 4(2) (3)
= 16 – 24
= -8 < 0 No real root
17. b 2 – 4ac = 0
⟹ (5m) 2 – 4(16) (1) =0
⟹ 25m 2 = 64

⟹ m2 =

⟹ m=
2
18. + 11x + 6 =0
2
⟹ + 9x + 2x + 6 =0
⟹ (x + 3 )=0
⟹ (x +

⟹ x=- & x= = -

19. 9x 2 – 3x – 2 = 0, for x = 2/3

9x 2 – 3x – 2 = 9 – 3( )–2

= 9x -2 -2 =4-4 =0

9x 2 – 3x – 2 = 0, x=-
 9x 2 – 3x – 2 = 9(- )² – 3 (- ) -2
= 9x + -2
= 1 + 1 – 2 = 2 -2= 0
Since the remainder in both cases is zero ,
.
20. Let the no be x
x – 20 = 69 ( )
2
⟹ x – 20 x = 69
⟹ x2 – 20 x – 69 = 0
⟹ x2 – 23 x + 3x – 69 = 0
⟹ x (x – 23) + 3 (x – 23) = 0
⟹ (x + 3) (x – 23) = 0
⟹ x = - 3 or x = 23

SECTION – C
21. 4s1 – 4s2 = 64
⟹ s1– ss = 64/4 =16
⟹ s1 = 16 + s2
⟹ (16 + s2)2 + (s2 )² = 640
⟹ 256 + s2 ² + 32s2 + s2 ² = 640
⟹ 2s2 ² + 32s2 – 640 +256 =0
⟹ 2s2 ² + 32s2 – 384 =0
⟹ s2 ² + 16s2 – 192 =0
⟹ s2 ² + 24s – 8s2 – 192 =0
⟹ s2 (s2 – 24) – 8 (s2+ 24) =0
⟹ (s2 + 24) (s2 – 8) =0
s2 = -24, 8 ( - 24 is not possible as side cannot be negative)
s2 = 8
s1 = 16 + 8 = 24
Sides of squares = 24m & 8m
22. x2 – (
⟹ x2 –
 ⟹ x2 – x= 0
b2 – 4ac = (-1)2 – 4(1) (0) = 1

x= =x= = = (1, 0)

23. 9x2−9(a+b)x+(2a2+5ab+2b2)=0
⟹9x2−9(a+b)x+ 2a2+4ab + ab + 2 b2 = 0
⟹9x2−9(a+b)x+[2a(a+2b)+b(a+2b)]=0
⟹9x2−9(a+b)x+[(2a+b)(a+2b)]=0
⟹9x2−3[(a+2b)(2a+b)]x+[(2a+b)(a+2b)]=0
⟹3x[3x−(a+2b)][3x−(2a+b)]=0
⟹3x=a+2b
⟹x= (a+2b)/3
or,
⟹3x=2a+b
⟹x = (2a+b)/3
x = [ (a+2b)/3 , (2a+b)/3 ]

24. Let the natural number be x

x² - 84 = 3 ( 8 + x)
x² - 84 = 24 + 3x
x² - 3x – 108 = 0

x = 24/2, -18/2
x = 12, - 9
Natural No. is 12

25. Let larger number be x

Let smaller number be y
x–y=5
x=5+y

- = = =
 ⟹ 24 [(y + 5) – y] = 5y (y + 5)

⟹ 24 [y + 5 – y] = 5y ² + 25y

⟹ 24 = y² +5y

y² + 5y – 24 = 0

y= =

y= = ,

y = 3, - 8

If smaller no is 3 then larger no is 8

If smaller no is -8 then larger no is -3.

SECTION – D
26. Let original length be ‘x’ m
Let original rate be Rs y / m
/x = y
(x + 5)(y – 2) = 200
⟹ xy – 2x + 5y – 10 = 200
⟹200 – 2x + 5y – 10 = 200
⟹-2x + 5y – 10 =0
⟹ -2x + 5 [ ] – 10 = 0

⟹-2x + – 10 x = 0

⟹ -2x² + 1000 – 10x = 0

⟹2x² + 10 x – 1000 = 0
⟹ x² + 5x – 500 = 0
⟹ x² + 25x – 20x – 500 = 0
⟹ x ( x + 25 ) – 20 [ x + 25] = 0
 ⟹ (x – 20)(x + 25) = 0

x = 20 or x = -25

27. Let ‘t1’ be the time taken to fill tank by one pipe.
Let ‘t2’be the time taken to fill by second pipe.
1
nd
Fraction of tank filled by 2 = 1 / t2
+ =

⟹ + =

⟹ =

⟹ =

⟹ 500 + 200t = 9t2 + 45t

⟹ 9t2 + 45t – 200t – 500 = 0
⟹ 9t2 – 155t – 500 = 0
⟹ 9t² - 180t + 25t – 500 = 0
⟹ 9t (t – 20) + 25 (t – 20) = 0
⟹ (t – 20) ( 9t + 25) = 0
⟹ t = 20 or t = -25/9 ( -25/9 is not possible as time cannot be negative)

28. Let numerator be ‘x’

+ =

⟹ x² + ( 2x + 1)² = ( 2x + 1)

⟹ 10[x2 + 4x2 + 4x + 1] = 29(2x2 + x)

 ⟹ 10[5x2 + 4x + 1] = 58x2 + 29
⟹ 50x2 + 40 x + 10 = 58x2 + 29 x
⟹

x= ,

x=2,-5/8
⸫ if Numerator
Then Denominator (2) + 1
⸫ Fraction = 2/5

29. At (t) minutes past 2 pm

According to given condition

(60 – t) = –3

⟹ 60 – t =

⟹ 240 – 4t = t² - 12
⟹ t² - 12 – 240 + 4t = 0
⟹ t² + 4t - 252 = 0
⟹ t² + 18t – 14t - 252 = 0
⟹ t ( t + 18 ) – 14 ( t + 18 ) = 0
⟹ ( t – 14 ) ( t + 18 ) = 0
⟹ t = 14 or t = - 18 (time cannot be negative)
30. =

⟹ =

⟹ abx = ( a + b + x ) ( bx + ax + ab)

⟹ abx = abx + a²x + a²b + b²x +abx + ab² + bx²+a x²+ abx

⟹ a²x + a²b + b²x +abx + ab² + bx²+a x²+ abx=0

⟹ a x²+ bx²+ a²x+ b²x+2abx+ a²b+ ab² = 0

⟹ x² ( a+ b) + x ( a² + b² + 2ab ) + a²b + ab² = 0

⟹ x² ( a+ b) + x (a + b) ²+ ab ( a + b)=0

⟹ ( a +b)[ x² + (a + b) x + ab ] = 0

⟹ x² + ( a +b ) x + ab = 0
⟹ ( x + a) ( x+ b) = 0
⟹ x = -a or x = – b

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