4 QUADRATIC EQUATIONS CHAPTER

4
CHAPTER
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QUADRATIC EQUATIONS
Points to Remember :
1. A quadratic equation in the variable x is of the form ax2 + bx + c = 0, where a, b, c are real numbers and
a  0.
2. A real number a is said to be a root of the quadratic equation ax2 + bx + c = 0, if a2 + b + c = 0. The
zeroes of the quadratic polynomial ax2 + bx + c and the roots of the quadratic equation ax2 + bx + c = 0
are the same.
3. If ax2 + bx + c, a  0 is factorisable into a product of two linear factors, then the roots of the quadratic
equation ax2 + bx + c = 0 can be found by equating each factor to zero.
4. The roots of a quadratic equation can also be found by using the method of completing the square.
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5. Quadratic Formula (Shreedharacharya’s rule) : The roots of a quadratic equation ax2  bx  c  0 are
given by
b  D
, where D = b2 – 4ac is known as discriminant.
2a
B
6. A quadratic equation ax2 + bx + c = 0 has
(i) Two distinct real roots, if b2 – 4ac > 0 i.e. D > 0
(ii) Two equal roots (i.e., coincident roots), if b2 – 4ac = 0 i.e. D = 0
(iii) No real roots, if b2 – 4ac < 0 i.e. D < 0.
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7.
x 2  a 2  x  a or x   a
8.
x2  a 2   a  x  a
ILLUSTRATIVE EXAMPLES
Example 1. Determine if x = 3 is a root of the given equation or not :
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Solution.
x 2  4 x  3  x 2  9  4 x 2  14 x  16
At x = 3
LHS =
2
2
x  4x  3  x  9
 (3) 2  4(3)  3  (3) 2  9  9  12  3  9  9  0  0  0
RHS  4 x2  14 x  16
 4(3) 2  14(3)  16  36  42  16  10
Since, LHS  RHS  3 is not a root of the given equation.
Example 2. Find the roots of the quadratic equation 8x2 – 22x – 21 = 0 by factorization.
Solution.
We have,


52
8 x2  22 x  21  0
8x2 – 28x + 6x – 21 = 0
4x (2x – 7) + 3 (2x – 7) = 0
QUADRATIC EQUATIONS
MATHEMATICS–X


(4x + 3) (2x – 7) = 0
4x + 3 = 0 or 2x – 7 = 0

x
Thus, x = –
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3
7
or x 
4
2
3
7
and x = are two roots of the quadratic equation 8 x2  22 x  21  0
4
2
Example 3. Find the roots of the quadratic equation 2 x 2  7 x  3  0 by the method of completing the
square.
Solution.
We have,
2x2  7 x  3  0

7
3
x2  x   0
2
2

7 3
x2  2  x    0
4 2

7 7 7 3
x2  2  x          0
4 4 4 2

7  49 3

x     0
4  16 2


7   49  24 

 x  
0
4   16 


7  25

0
x  
4  16


7
25  5 

  
x  
4  16  4 


7
5
x  
4
4

x
7 5
7
5
 or x   
4 4
4
4

x
5 7
5 7
 or x 

4 4
4 4

x  3 or x 
2
2
2
2
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2
2
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Thus, x = 3 and x =
MATHEMATICS–X
2
B
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1
2
1
are two roots of the quadratic equation 2 x 2  7 x  3  0 .
2
QUADRATIC EQUATIONS
53
Example 4. Using quadratic formula, solve the following equation for x :
abx 2  (b 2  ac ) x  bc  0
Solution. We have, abx2 + (b2 – ac) x – bc = 0
Comparing this equation with Ax2 + Bx + C = 0, we get
A  ab, B  b 2  ac, C  bc
 D  B 2  4 AC
 (b 2  ac ) 2  4(ab )(bc )
 b4  a 2 c 2  2ab2 c  4ab 2 c
 b4  2ab 2 c  a 2 c2
 (b 2 ) 2  2(b 2 )( ac )  ( ac ) 2
 (b 2  ac) 2  0
So, the roots of the given equation are real and are given by :
B  D
2A
(b 2  ac )  (b2  ac )2
x
2(ab)
x

[CBSE 2005]
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(b2  ac )  (b2  ac )
2ab
(b2  ac)  (b 2  ac )
(b 2  ac)  (b 2  ac)
 x
,x
2ab
2ab
2ac
2b 2
,x
 x
2ab
2ab
c
b
 x ,x
b
a
c
b
Thus, x  and x 
are two roots of the quadratic equation abx2 + (b2 – ac) x – bc = 0.
b
a
Example 5. Find the value(s) of k for which the following equation has equal roots.

x
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( k – 12) x 2 + 2 (k – 12 ) x + 2 = 0
Solution.
B
We have, ( k  12) x 2  2( k  12) x  2  0
Here, a = k – 12, b = 2 (k – 12), c = 2
 D = b2 – 4ac = 4(k – 12)2 – 4 (k – 12) × 2
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 4 (k  12) [(k  12)  2]
 4(k  12)(k  14)
The given equation will have equal roots if D = 0.

4 (k  12) (k  14)  0
 k – 12 = 0 or k – 14 = 0
 k = 12 or k = 14 Ans.
Example 6. 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.
[CBSE 2006 C]
Solution. Let the tens digit be x.
Then, the units digit  18
x
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QUADRATIC EQUATIONS
MATHEMATICS–X
 Number = 10 (tens digit) + unit digit  10x 
18
x
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and, number obtained by interchanging the digits  10  18  x
x
18 
18



According to given question,  10 x    63   10   x 
x
x




18  
18


 10 x    10   x   63
x 
x



10 x 

9x2 – 63x – 162 = 0

x2 – 7x – 18 = 0

(x – 9) (x + 2) = 0

x – 9 = 0 or x + 2 = 0

x = 9 or x = – 2
18 180

 x  63
x
x
But, a digit can never be negative. So, x = 9.
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Hence the required number  10 x  18  10(9)  18 = 90 + 2 = 92 Ans.
x
9
Example 7. The sum of the ages of a woman and her daughter is 40 years. The product of their ages five years
ago was 125 years. Find their present ages.
Solution.
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Let the present age of the woman be x years. Then, the present age of her daughter is (40 – x)
years.
Five years ago, Age of woman = (x – 5) years
and, Age of daughter = (40 – x – 5) years = (35 – x) years.
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According to the given question, (x – 5) (35 – x) = 125

35 x – x2 – 175 + 5x – 125 = 0

x2 – 40 x + 300 = 0

(x – 30) (x – 10) = 0

x – 30 = 0 or x – 10 = 0

x = 30 or x = 10
when, x = 30, then 40 – x = 40 – 30 = 10
Thus, when the woman’s present age is 30 years, her daughter’s present age is 10 years.
When, x = 10, then 40 – x = 40 – 10 = 30, which is absurd, because the woman’s age cannot be less
than her daughter’s age.
Hence, present age of woman = 30 years
and, present age of daughter = 10 years.
MATHEMATICS–X
QUADRATIC EQUATIONS
55
Example 8. A plane left 30 minutes later than the schedule time and in order to reach its destination 1500
km away in time it has to increase its speed by 250 km/hr from its usual speed. Find its usual
speed.
Solution. Let the usual speed of the plane be x km/hr. Then
Time taken to cover 1500 km with the usual speed 
Time taken to cover 1500 km with the speed of (x + 250) km/hr  1500 hrs.
x  250
1500 1500
30
According to given question,


x
x  250 60
1500 x  1500  250  1500 x 1


x( x  250)
2

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1500
hrs.
x
1500  250 1

x 2  250 x 2

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x2  250 x  750000  0
 x2 + 1000x – 750 x – 750000 = 0
 x (x + 1000) – 750 (x + 1000) = 0
 (x – 750) (x + 1000) = 0
 x = 750 or x = – 1000
 x = 750
[ speed cannot be negative]
Hence, the usual speed of the plane is 750 km/hr.
B
Example 9. The hypotenuse of a right triangle is 3 5 cm. If the smaller side is tripled and the larger side is
doubled, the new hypotenuse will be 15 cm. Find the length of each side.
Solution. Let the smaller side of the right triangle be x cm and the larger side be y cm. Then,
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x2  y 2  (3 5)2
[ using Pythagoras theorem]
 x2 + y2 = 45
...(1)
If the smaller side is tripled and the larger side be doubled, the new hypotenuse is 15 cm.
 (3x)2 + (2y)2 = (15)2
 9x2 + 4y2 = 225
...(2)
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from eqn. (1), we get y 2  45  x 2 . Putting this value of y2 in eqn. (2), we get
9x2 + 4 (45 – x2) = 225
 5x2 + 180 = 225  5x2 = 45  x2 = 9  x = ± 3
But, length of a side cannot be negative.  x = 3
Putting x = 3 in eqn. (1), we get
9 + y2 = 45  y2 = 36  y = ± 6  y = 6
[ y cannot be negative]
Hence, the length of the smaller side is 3 cm and the length of the larger side is 6 cm.
Example 10. By increasing the list price of a book by Rs. 10 a person can buy 10 less books for Rs. 1,200. Find
the original list price of the book.
[CBSE 2007]
Solution. Let the original list price of book = Rs. x.
and, the increased list price of a book = Rs. (x + 10)
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QUADRATIC EQUATIONS
MATHEMATICS–X
Total number of books which can be bought for Rs. 1200 
1200
x
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Total number of books which can be bought in new case  1200
x  10
According to given question,

1200 1200

 10
x
x  10
1200 ( x  10  x )
 10
x ( x  10)
 12000 = 10 x (x + 10)
 x2 + 10x – 1200 = 0
 x2 + 40x – 30x – 1200 = 0
 x (x + 40) – 30 (x + 40) = 0
 (x – 30) (x + 40) = 0
 x – 30 = 0 or x + 40 = 0
 x = 30 or x = – 40
 x = 30
[ x cannot be negative]
Thus, the original list price of the book = Rs. 30 Ans.
B
PRACTICE EXERCISE
Section - A
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1. Which of the following are quadratic equations ?
(a) x2 + 7x – 12 = 0
(d) 7x2 = 0
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(b) 9x2 – 12 x = 0
2
(e) x 
1
4
x2
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(c) x 
4
 x3
x
(f) x( x  7)  x 2  4 x  2
2. In each of the following, determine whether the given values are the solutions of the given quadratic
equations or not :
(b) 7x2 + 4x – 20 = 0; x = – 2 , x 
(a) x2 – 5x + 6 = 0 ; x = 2, x = 3
(c) x 
(e)
1 25
2
3
 ; x ,x
x 12
3
4
3x 2  11x  6 3  0; x  
10
7
(d) x 2  2 3  9  0 ; x   3, x  3 3
2
3
, x  3 3
3. In each of the following, find the value of p for which the given value is a solution of the given equation:
(a) (2 p  1) x 2  2 x  3  0; x  2
(b) x 2  2 ax  2 p  0; x   a
(c) x 2  x(a  b)  p  0; x  b
(d) px2  2 x  4  0; x  2
MATHEMATICS–X
QUADRATIC EQUATIONS
57
4. If x = 5 and x  
5
are the roots of the equation ax2 – 15x + b = 0. Find the value of a and b.
4
5. Determine whether x 
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1
3
and x  are solutions of the equation 2 x 2  5 x  3  0 or not.
2
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Solve the following equations by the method of factorization (6-20)
6. 3x2 – 5x + 2 = 0
7. 25x2 – 10 x – 8 = 0
8. x2 – 3x – 28 = 0
9. 3x2 + 12ax – a2 = 0, where a is real.
10. 5x2 + 16x = – 12
12.
14.
11. 6 2 x 2  5x  3 2  0
2
2
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13. x2  ( 3  1) x  3  0
6a 2 x 2  7abx  3b 2  0
2
2
2
4
4
15. 4 x  4a x  (a  b )  0
x  2ax  a  b  0
16. a2b2x2 + b2x – a2x +1 = 0
[CBSE 2005] 17.
18.
x  3 1  x 17


x2
x
4
20.
x 1 x  3
1

 3 ; x  2, 4 [CBSE 2005]
x2 x4
3
19.
[CBSE 2004]
1
1 1 1
   , ab  0
ab x a b x
x  1 2x  1 5
1

 ;x
,1
2x  1 x  1 2
2
B
[CBSE 2005]
Solve the following quadratic equations (if possible) by the method of completing the square (21-30)
22. 2x2 + x – 4 = 0
x2  8 x  0
23. 9x2 – 15x + 6 = 0
21.
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25. 4x2 + 4 3 x + 3 = 0
27. 15x2 – 28 = x
29.
24. 4x2 + 3x + 9 = 0
x2  ( 2  1) x  2  0
26. 8x 2  10 x  3
28. x 2  6 2 x  10  0
30. a 2 x2  3abx  2b 2  0
Determine the discriminant of each of the following quadratic equations (31-35)
31. 3x2 – 5x +2 = 0
32. 5x2 + 4x – 1 = 0
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33. x2 + 2ax + 3b = 0
34.
2 x 2  3x  3 2  0
35. –4x2 = 6 – 3x
Solve the following quadratic equations using Quadratic Formula (36-50)
36. x2 – 9x + 20 = 0
37. x2 – 6x + 9 = 0
38. x2 – 2xb + b2 = 0
40. 3x2 + 2 5x  5 = 0
2
2
3
39. x2 – 4x + 4 =  
2
2
41. 5 x  9 x  14 5  0
42.
2 x  (q  2 p ) x  pq  0
43. 3 y 2  (6  4 a ) y  8a  0
44.
a
b

 2; x  a , b
xb xa
45. 4x2 – 2 (a2 + b2) x + a2b2 = 0
[CBSE 2004]
47. 9x2 – 9(a + b)x + (2a2 + 5ab + 2b2) = 0
[CBSE 2004]
46. abx2 – (a + b)2 (x – 1) = 0
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QUADRATIC EQUATIONS
MATHEMATICS–X
48.
abx 2  (b 2  ac ) x  bc  0 [CBSE 2005]
50.
4
5
3
3 
; x  0,
[CBSE 2006 C]
x
2x  3
2
49. x 2  x (a  2)  (a  1)  0
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Solve the following Quadratic Equations (51-55)
51.
p 2 x 2  ( p 2  q 2 ) x  q 2  0 [CBSE 2004]
52.
1
2
4


; x  1,  2,  4
x 1 x  2 x  4
53.
6 x 2  (12  8a ) x  16a  0
54.
2 x  3 2 x  7 16

 ; x  2, 4
x2
x4
3
55.
xa x b a b

  ; x  a, b
xb xa b a
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Determine the nature of the roots of the following quadratic equations (56-60)
56. 9x2 – 6x + 1 = 0
57. x2 + x – 12 = 0
59. 4x2 – 4x + 1 = 0
4 3x 2  5x  2 3  0
60. (x – 2a) (x – 2b) = 4ab
Find the value(s) of k for which the roots of following quadratic equations are real and equal (61-65)
58.
61.
B
62. kx2 – 2 5 x + 4 = 0
2 x 2  10 x  k  0
64. k 2 x 2  2(2k  1) x  4  0
63. (k + 4)x2 + (k + 1)x + 1 = 0
65. (k + 1)x2 – 2 (k – 1) x + 1 = 0
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[CBSE 2001(C)]
[CBSE 2002 (C)]
Find the value(s) of p for which each of the following equations has real roots (66-70)
66. 5x2 – 6x + p = 0
67. 9x2 – 5 px + 1 = 0
68. x2 + 4x + p = 0
69. 2x2 + px + 2 = 0
70. 4x2 – 3px + 1 = 0
71. For what value of k, (4 – k) x2+ (2k + 4)x + (8k + 1) = 0, is a perfect square.
ac
.
2
[CBSE 2002 (C)]
72. If the roots of the equation (b – c)x2 + (c – a) x + (a – b) = 0 are equal, then prove that b 
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73. Find the values of k so that the equation (k + 2) x2 – (7 – k) x + 3 = 0 has :
(a) equal roots
(b) distinct roots
(c) no real roots.
74. If the equation (1 + m2) x2 + 2mcx + (c2 – a2) = 0 has equal roots, prove that c 2  a 2 (1  m2 ).
75. Prove that both the roots of the equation (x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0 are real and
equal only when a = b = c.
Section - B (Word Problems)
Problem Related to Numbers (76–85)
76. The sum of two numbers is 16. The sum of their reciprocals is
1
. Find the numbers.
3
77. The sum of the squares of two consecutive odd positive integers is 290. Find them.
MATHEMATICS–X
QUADRATIC EQUATIONS
59
78. Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller
part by 164.
79. Find two consecutive numbers whose squares have the sum 85.
80. If the sum of first n even natural numbers is 420, find the value of n.
81. Two numbers differ by 3 and their product is 504. Find the numbers.
[CBSE 2002(C)]
82. A two digit number is such that the product of the digits is 14. When 45 is added to the number, then the
digits are reversed. Find the number.
83. A two digit number is such that the product of its digit is 10. When 27 is subtracted from the number, the
digits interchange their places. Find the number.
84. The denominator of a fraction is 5 more than its numerator. The sum of the fraction and its reciprocal is
3
11
. Find the fraction.
14
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85. 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.
Problems related to time, Distance and Speed (86–92)
86. The speed of a boat in still water is 8 km/hr. It can go 15 km upstream 22 km downstream in 5 hours. Find
the speed of the stream.
87. A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km per hour more, it would
have taken 30 minutes less for the journey. Find the original speed of the train.
[CBSE 2006 (C)]
88. A passenger train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/hr from its
usual speed. Find the usual speed of the train.
89. Two trains leave a railway station at the same time. The first train travels due west and the second train
due north. The first train travels 5 km/hr faster than the second train. If, after 2 hours, they are 50 km
apart, find the speed of each trains.
90. In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was
reduced by 200 km/hr and the time of flight increased by 30 minutes. Find the duration of the flight.
91. A faster train takes one hour less than a slow train for a journey of 200 km. If the speed of the slow train
is 10 km/hr less than that of the fast train, find the speed of the two trains.
92. The speed of a boat in still water is 15 km/hr. It can go 30 km upstream and return downstream to the
original point in 4 hour 30 minutes. Find the speed of the stream.
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B
Problems related to Ages (93–95)
93. One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his sons age.
Find their present ages.
94. The sum of the ages of a man and his son is 40 years. The product of their ages is 144 years. Find their
present ages.
95. 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.
[CBSE 2006 C]
Problems related to Geometry (96–102)
96. Two circles touch externally. The sum of their areas is 130  sq.cm and the distance between their centres
is 14 cm. Find the radii of the circles.
97. The length of a rectangular ground is greater than twice its breadth by 5 m. The area of the ground is 1125
sq. m. Find the length and breadth of the ground.
60
QUADRATIC EQUATIONS
MATHEMATICS–X
98. The sides of a right angled triangle containing the right angle are 5x cm and (3x – 1) cm. If the area of the
triangle be 60 sq. cm., calculate the length of the sides of the triangle.
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99. 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 th area of the larger square is added to three times the area of the smaller square, the
result is 203 cm2. Find the sides of the squares.
100. The perimeter of a right triangle is 60 cm. Its hypotenuse is 25 cm. Find the area of the triangle.
101. A carpet is placed in a room 6m × 4m, leaving a border of a uniform width all around it. Find the width of
the border, if the area of the carpet is 8 sq. m.
102. The length of the hypotenuse of a right-angled triangle exceeds the base by 1 cm and also exceeds twice
the length of the altitude by 3 cm. Find the length of each side of the triangle.
[CBSE 2003]
Problems related to Cost, Expenditure (103–108)
J
A
103. If the list price of a toy is reduced by Rs. 2, a person can buy 2 toys more for Rs. 360. Find the original
price of the toy.
[CBSE 2002(C)]
104. A person on tour has Rs. 360 for his expenses. If he extends his tour for 4 days, he has to cut down his
daily expenses by Rs. 3. Find the original duration of the tour.
105. Rs. 6500 were divided among a certain number of persons. Had there been 15 more persons, each would
have got Rs. 30 less. Find the original number of the persons.
B
106. Some students planned a picnic. The budget the food was Rs. 500. But, 5 of them failed to go and thus
the cost of food for each member increased by Rs. 5. How many students attended the picnic?
107. A shopkeeper buys a number of books for Rs. 80. If he had bought 4 more books for the same amount,
each book would have cost Rs. 1 less. How many books did he buy?
108. A piece of cloth costs Rs. 200. If the piece was 5 m longer and each metre of cloth costs Rs. 2 less, the
cost of the piece would have remained unchanged. How long is the piece and what is the original rate per
metre?
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I
Miscellaneous Questions (109–125)
109. A takes 15 days less than the time taken by B to finish piece of work. If both A and B together can finish
the work in 18 days, find the time taken by B to finish the work.
1
minutes. If one pipe takes 3 minutes more than the
13
other to fill it, find the time in which each pipe would fill the cistern.
111. In a class test, the sum of Nishu’s marks in Mathematics and Science is 30. Had she got 2 marks more in
Mathematics and 3 marks less in Science, the product of their marks would have been 210. Find her marks
in the two subjects.
110. Two pipes running together can fill a cistern in 3
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A
n(n  3)
diagonals. How many sides has a polygon with 54 diagonals?
2
113. In an auditorium, the number of rows was equal to number of seats in each row. When the number of
rows was doubled and the number of seats in each row was reduced by 10, the total number of seats
increased by 300. How many rows were there?
112. A polygon of n side has
7
times the square root of the number are playing on the shore of a tank.
2
The two remaining ones are playing with amorous fight, in the water. What is the total number of swans?
115. A line AB is 8 cm in length. AB is produced to P such that BP2 = AB. AP. Find the length of BP.
114. O girl! Out of a group of swans,
MATHEMATICS–X
QUADRATIC EQUATIONS
61
116. The difference of the squares of two numbers is 45. The square of the smaller number is four times the
larger number. Find the numbers.
117. The length of a rectangle exceeds its breadth by 5 m. If the breadth were doubled and the length reduced
by 9 m, the area of the rectangle would have increased by 140 sq. m. Find its dimensions.
118. A farmer wishes to start a 300 sq. m rectangular vegetable garden. Since he has only 50 m barbed wire, he
fences three sides of the rectangular garden, letting his house compound wall act as the fourth side
fence. Find the dimensions of the garden.
119. In an auditorium, seats are arranged in rows and columns. The number of rows was equal to the number
of seats in each row. When the number of rows was doubled and the number of seats in each row is
reduced by 10, the total number of seats increased by 300. Find :
(i) the number of rows in the original arrangement.
(ii) the number of seats in the auditorium after re-arrangement.
120. A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 sq. m
more than the area of a park that was already been made in the shape of an isosceles triangle with its base
as the breadth of the rectangular park and of altitude 12 m. Find its length and breadth.
12 m
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B
J
A
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121. Sum of the areas of two squares is 468 m2. If the difference of their perimeter is 24 m, find the sides of the
two squares.
122. X and Y are centres of circle of radius 9 cm and 2 cm and XY = 17 cm. Z is the centre of a circle of radius
r cm, which touches the above circles externally. Given that XZY = 90°, find the value of r.
123. The area of an isosceles triangle is 60 cm2 and the length of each of its equal sides is 13 cm. Find its base.
124. The angry Arjun carried some arrows for fighting with Bheesm. With half the arrows, he cut down the
arrows thrown by Bheesm on him and with six other arrows, he killed the rath driver of the Bheesm. With
one arrow each, he knocked down respectively the rath, flag and the bow of Bheesm. Finally, with one
more than four times the square root of the total number of arrows, he laid Bheesm unconscious on an
arrow bed. Find the total number of arrows Arjun had.
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125. Out of a number of Saras birds, one fourth the number are moving about in lotus plants;
1
th coupled
9
1
th as well as 7 times the square root of the number move on a hill; 56 birds remain in
4
vakula trees. What is the total number of birds?
(along) with
HINTS TO SELECTED QUESTIONS
15. 4x2 – 4a2 x + (a4 – b4) = 0

4x2 – [2(a2 + b2) + 2 (a2 – b2)] x + (a2 – b2) (a2 + b2) = 0

4x2 – 2(a2 + b2) x – 2 (a2 – b2) x + (a2 – b2) (a2 + b2) = 0

2x [2x – (a2 + b2)] – (a2 – b2) [2x – (a2 + b2)] = 0
62
QUADRATIC EQUATIONS
MATHEMATICS–X
17.

[2x – (a2 – b2)] . [2x – (a2 + b2)] = 0

x
a2  b2
a 2  b2
or x 
2
2

1
1 1 1
  
ab x x a b

x  ( a  b  x) a  b

x( a  b  x )
ab
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A

x (a + b + x) = ab  x2 + ax + bx + ab = 0

x (x + a) + b (x + a) = 0  (x + a) (x + b) = 0  x = – a or x = – b.
46. abx2 – (a + b)2 (x – 1) = 0  abx2 – (a + b)2 x + (a + b)2 = 0
here, A = ab, B = – (a + b)2 and C = (a + b)2. Now use quadratic formula.
47.
9 x 2  9( a  b ) x  (2 a 2  5ab  2b 2 )  0
here, D  [ 9(a  b )]2  4(9)(2a 2  5ab  2b 2 )
B
 81 ( a 2  b 2  2ab)  36(2 a 2  5ab  2b 2 )
 9( a  b) 2  0

54.
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A
1
1 1 1
  
a b  x a b x
x
 B  D 9(a  b)  3(a  b) 2a  b a  2b


,
2A
2(9)
3
3
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2 x  3 2 x  7 16


x2
x4
3

2 x  4  1 2 x  8  1 16


x2
x4
3
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A

2( x  2)  1 2( x  4)  1 16


x2
x4
3
 2
1
1
16
2

x2
x4 3
1
1
4

 .
x2 x4 3
Now take LCM and simplify to get quadratic equation.

71. For perfect square, d = 0  (2k + 4)2 – 4(4 – k) (8k + 1) = 0.
72. For equal roots, d = 0  (c – a)2 – 4(b – c) (a – b) = 0
 c2 + a2 – 2ca – 4 (ab – b2 – ca + bc) = 0
 (–2b + a + c)2 = 0
 –2b + a + c = 0  a + c = 2b.
MATHEMATICS–X
QUADRATIC EQUATIONS
63
74. For equal roots, d = 0

(2mc)2 – 4 (1 + m2) (c2 – a2) = 0

4m2c2 – 4 (c2 – a2 + m2c2 – a2m2) = 0

m2c2 – c2 + a2 – m2c2 + a2m2 = 0

c2 = a2 (1 + m2).
75. The given equation can be written as
3x2 – 2x(a + b + c) + (ab + bc + ca) = 0

D = 4(a + b + c)2 – 12 (ab + bc + ca)
= 4 (a2 + b2 + c2 – ab – bc – ca)
= 2 [(a – b)2 + (b – c)2 + (c – a)2]  0
[ (a – b)2  0, (b – c)2  0 and (c – a)2  0]
This shows that both the roots are real.
For equal roots, D = 0

(a – b)2 + (b – c)2 + (c – a)2 = 0

a – b = 0, b – c = 0, c – a = 0

a = b = c.
85. Let total number of camels be x.
Then,


Put
x
 2 x  15  x
4
B
x
3x
 15  2 x 
 15  2 x  3x  60  8 x
4
4
3 x  8 x  60  0
x  y, we get 3y2 – 8y – 60 = 0
x
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Solving, we get y = 6 or y 
10
3
2
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when, y  10  x   10   100 . But number of camels cannot be a fraction.
3
9
 3 
When, y = 6  x = (6)2 = 36.
95. Seven years ago, let Swati age be x years. Then, seven years ago Varun’s age was 5x2 years.
 Swati’s and Varun’s present ages are (x + 7) years and (5x2 + 7) years, respectively.
Three years hence, Swati’s and Varun’s ages will be (x + 10) years and (5x2 + 10) years respectively.
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2
(5 x 2  10)
5
3

2x2 – x – 6 = 0  x = 2,  . But x cannot be negative.
2
 Swati’s present age = (2 + 7) years = 9 years
Varun’s present age = (5 × 22 + 7) years = 27 years.
96. Let radii of circles be r1 cm and r2 cm.
according to question, x  10 
Then, r12  r22  130   r12  r22  130
...(1)
also, r1  r2  14
...(2)
Solve (1) and (2) together now.
64
QUADRATIC EQUATIONS
MATHEMATICS–X
102. Let base and altitude of right-angled triangle be x cm and y cm.
Then, hypotenuse = x + 1 = 2y + 3
...(1)
2
2
2
Also, x + y = (x + 1)
...(2)
Solve (1) and (2) together now.
109. Suppose B alone takes x days to finish the work.
1
1
1
 .
Then, 
x x  15 18
110. Suppose the faster pipe takes x minutes to fill the cistern. Then, slower pipe will take (x + 3) minutes.
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According to question, 1  1  13
x x  3 40
118. Let the length of one side be x metres and other side by y metres.
Then, x + y + x = 50  2x + y = 50.
...(1)
Also, xy = 300
...(2)
Solve (1) and (2) together now.
120. Let l and b be the length and breadth of the park.
Then, b = l – 3
...(1)
According to given question,
1
 b 12  4  l  b
2
 6b + 4 = l × b
Solve (1) and (2) together now.
124. Suppose Arjun had x arrows.
Then,
x
 6  3  4 x 1  x
2

x  8 x  20  0
B
...(2)
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Let
x  y. Then, y 2  8 y  20  0  y = 10 or –2.

y = 10

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[ y cannot be negative]
x  10  x  100.
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MULTIPLE CHOICE QUESTIONS
Mark the correct alternative in each of the following :
1. If one root of 2x2 + kx – 6 = 0 is 6, the value of k is :
(a) –11
(b) 11
(c) 2
2. The roots of quadratic equation
7
3x 2  10 x  7 3  0 are :
7
(c)  3 and 7
3
3
3. The value(s) of x satisfying the equation a2x2 – 3abx + 2b2 = 0 is :
(a)  3 and
(a)
3
2b b
,
a a
MATHEMATICS–X
(b) 3 and
(b)
2b b
,
a
a
(d) 3
(c)
2b b
,
a a
QUADRATIC EQUATIONS
(d) none of these
(d) none of these
65
4. The discriminant of the quadratic equation 3 x2  8 x  3  0 is :
(a) –100
(b) 53
(c) 28
(d) none of these
2
5. The value(s) of k for which the equation (k + 1) x – 2 (k – 1) x + 1 = 0 has real and equal roots is:
(a) k = 0, –3
(b) k = 0, 3
(c) k = 0, 4
(d) none of these
2
6. The value(s) of k for which the equation 9x + 3kx + 4 = 0 has real roots is :
(a) k  4 or k  4 (b) k  4 or k  4
(c) k  2 or k  2
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(d) k  2 or k  2
7. The value of 12  12  12  ..... is :
(a) –2
(b) –4
(c) 2
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8. If the equation x2  kx  1  0 does not possess real roots, then :
(a) –3 < k < 3
(b) –2 < k < 2
(c) k > 2
9. The value of k for which the equation :
(4  k ) x 2  (2 k  4) x  (8k  1)  0 is a perfect square, is :
(d) 4
(d) k < – 2
(a) k = 0, 3
(b) k = 0, –3
(c) k = 3, –3
(d) none of these
2
10. The value of m for which the roots of the equation (3m + 1) x + (m + 11) x + 9 = 0 are equal, is :
(a) m = –1, m = 85 (b) m = – 1, m = – 85
(c) m = 1, m = 85
(d) none of these
B
VERY SHORT ANSWER TYPE QUESTIONS (1 MARK QUESTIONS)
1. Check whether x (x + 3) + 8x = (x + 5) (x – 5) is a quadratic equation.
2. If x 
3
is the root of the quadratic equation 2x2 – kx + 3 = 0, then find the value of k.
2
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3. What are the roots of the equation 2x2 + 3x – 2 = 0?
4. Write the quadratic equation for the following :
The sum of two numbers is 27 and their product is 182.
5. Represent the following in the form of a quadratic equation :
The sum of the ages of father and his son is 45 years. Five years ago, the product of their ages (in years)
was 124.
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6. Determine whether x 
1
2
and x  are the roots of 9x2 – 3x – 2 = 0.
3
3
7. Form a quadratic equation whose roots are –2 and 3.
8. Form a quadratic equation whose sum of roots is –3 and product of roots is 5.
9. If  and  are the roots of the equation 3x2 – 9x + 7 = 0, then find  +  + .
10. For what value of k, x = a is a solution of the equation x 2  (a  b ) x  k  0 .
11. If one root of the equation 3x2 + px + 4 = 0 is
12. If one root of a quadratic equation is
66
2
, find the value of p.
3
3 2
, then what is the other root?
4
QUADRATIC EQUATIONS
MATHEMATICS–X
13. Write the discriminant of the equation 6a2x2 – 7abx – 3b2 = 0.
14. Write the nature of the roots of the equation 3 x 2  4 3x  1  0.
J
A
15. For what value of k, the equation 3x2 + kx + 4 = 0 have equal roots?
16. For what value of k, the equation kx2 + 6x + 1 = 0 have real roots?
17. Solve the quadratic equation : (x + 3)2 – 16 = 0.
10
, what is the number?
3
19. Find the ratio of sum and product of the roots of the equation 3x2 – 8x + 12 = 0.
20. The sum of squares of two consecutive positive integers is 365. Express this in the form of an quadratic
equation.
18. If sum of a whole number and its reciprocal is
PRACTICE TEST
M.M : 30
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Time : 1 hour
General Instructions :
Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.
B
1. Divide 26 into two parts, whose sum of squares is 346. Frame an equation for the given statement.
2. Solve for x : 4x2 – 4ax + a2 – b2 = 0.
3. Find the values of k so that the quadratic equation x2 – 2x (1 + 3k) + 7 (3 + 2k) = 0 has equal roots.
4. Solve the equation 2x2 – 5x + 3 = 0 by the method of completing the square.
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5. The sum of two numbers is 16 and the sum of their reciprocals is
6. Solve for x using quadratic formula :
a 2b 2 x 2  b 2 x  a 2 x  1  0
1
. Find the numbers.
3
7. Find the values of k for which the equation x2 + 5kx + 16 = 0 has no real roots.
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8. Solve for x :
1
1
11


; x  4, 7
x  4 x  7 30
9. A plane left 40 minutes late due to bad weather and in order to reach its destination, 1600 km away in time,
it had to increase its speed by 400 km/hr from its usual speed. Find the usual speed of the plane.
10. A two digit number is four times the sum of its digits and twice the product of its digits. Find the number.
ANSWERS OF PRACTICE EXERCISE
1. (a), (b), (d)
2. (a) x = 2 and x = 3 are solutions.
(b) x = – 2 is a solution, but x  
(c) x 
10
is not a solution.
7
2
3
is not a solution, x  is a solution.
3
4
MATHEMATICS–X
QUADRATIC EQUATIONS
67
(d) x   3 and x  3 3 are solutions.
(e) x  
3. (a) p  
2
3
5
8
4. a = 4, b = – 25
6. 1,
2
3
(b) p  
(c) p  ab
4 2
,
5 5
2 3
,
3 2 2
12.
b 3b
,
3a 2a
14. a – b, a + b
15.
a 2  b 2 a 2  b2
,
2
2
16.
1 1
,
a2 b2
18.
22.
1  33 1  33
,
4
4
25.
 3
3
,
2
2
19. –1
2, 1
33. 4a2 – 12b
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14
5,
45.
a2 b2
,
2 2
5
49. –1, –(1 + a)
53.
68
2,
4a
3
23. 1,
2
3
7
4
,
5 3
13. 1, 3
17. –a, –b
21. 0, 8
24. No real roots
3 1
,
4 2
27.
30.
2b b
,
a a
31. 1
32. 36
35. –87
36. 4, 5
34. 33
38. b, b
41.
5
2
a
3
26.
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37. 3, 3
B
20. 5,
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9. a ,
8. 7, –4
11.
2
4,
9
(d) p = 3
1
3
is not a solution, but x  is a solution.
2
2
5. x 
7.
a2
2
10. –2,  6
5
29.
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A
and x  3 3 are solutoins.
42. p,
46.
q
2
ab ab
,
a
a
50. –2, 1
54. 5,
5
2
39.
7 1
,
2 2
43. 2,
47.
4 a
3
2a  b a  2b
,
3
3
51. 1,
q2
p2
55. 0, a  b
QUADRATIC EQUATIONS
28.
2,5 2
40.  5,
5
3
44. a  b,
a b
2
48.
b c
,
a b
52. 2  2 3
56. Real and equal
MATHEMATICS–X
57. Real and unequal
61.
k
25
2
65. k = 0 or 3
69.
p  4 or p  4
58. no real roots
59. Real and equal
60. Real and unequal
64. k 
62. k 
5
4
63. k = – 3 or 5
66. p 
9
5
67. p 
70. p  
4
4
or p 
3
3
6
6
or p  
5
5
76. 4, 12
78. 10, 6
79. 6, 7
80. n = 20
82. 27
83. 52
84.
86. 3 km/hr
87. 45 km/hr
88. 25 km/hr
90. 1 hour
91. 40 km/hr, 50 km/hr
92. 5 km/hr
94. 36 years and 4 years
2
7
B
95. Swati’s present age = 9 years, varun’s present age = 27 years
45
m
2
96. 11 cm and 3 cm
97. length = 50 m, breadth 
99. 5 cm and 8 cm
100. 150 cm2
101. 1 m
103. Rs. 20
104. 20 days
105. 50
107. 16
108. 20 m, Rs. 10
109. 45 days
111.
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68. p  4
71. k  0, 3
73. (a) k = 1, 25 (b) k > 25 or k < 1 (c) 1 < k < 25
110. 5 minutes and 8 minutes
1
4
J
A
77. 11, 13
81. 21, 24 or – 24, – 21
85. 36
89. 20 km/hr, 15 km/hr
93. 7 years and 49 years
98. 15 cm, 8 cm, 17 cm
102. 5 cm, 12 cm, 13 cm
106. 20
Marks in mathematics = 13, Marks in science = 17
M
A
OR
Marks in Mathematics = 12, Marks in science = 18
112. 12
113. 30 rows
114. 16
115. 4(1  5) cm
116. 6 and 9
117. 25 m × 20 m
118. 20 m × 15 m or 30 m × 10 m
119. (i) 30 (ii) 1200
120. 7 m, 4 m
121. 18 m and 12 m
123. 10 cm or 24 cm
124. 100
125. 576
122. r = 6 cm
ANSWERS OF MULTIPLE CHOICE QUESTIONS
1. (a)
6. (a)
2. (c)
7. (d)
MATHEMATICS–X
3. (a)
8. (b)
4. (c)
9. (a)
QUADRATIC EQUATIONS
5. (b)
10. (c)
69
ANSWERS OF VERY SHORT ANSWER TYPE QUESTIONS
2. k = 5
3. 2,
x2  45 x  324  0
6. yes
1. No
5.
8. x2 + 3x + 5 = 0
9.
1
2
16
3
2
4. x  27 x  182  0
10. k = ab
3 2
4
13.
121a2b2
14. Real and distinct roots
15. k  4 3
16.
k 9
17. 1, –7
18. 3
19. 2 : 3
11.
p  8
12.
20. x2 + x – 182 = 0
J
A
ANSWERS OF PRACTICE TEST
1. x2 + (26 – x)2 = 346
3
2
8
8
7.   k 
5
5
10. 36
2.
4. 1,
8. 1, 2
T
I
M
A
70
a b
ab
,x
2
2
5. 4, 12
J
A
7. x2  x  6  0
B
10
9
1
1
6.  2 , 2
a b
3. 2 or
9. 800 km/hr
QUADRATIC EQUATIONS
MATHEMATICS–X