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CBSE Practice papersQUADRATIC EQUATIONS(Answers and Solutions)class-10th-Mathematics Number of Questions: 136For Questions, Go to www.micromerits.com -> Select your grade ->Select Learn ->Select SubjectPage link - http://www.micromerits.com

1: Option 2Solution Details

The general form of a quadratic equation is ax2 + bx + c = 0; a ≠ 0

Here, for 1) x2 + 7x = x2 + 9 ⇒ 7x − 9 = 0, no x2 term, hence not a quadratic equation

2) 2x2 + 7x = x − 1 ⇒ 2x2 + 6x + 1 = 0, this satisfies the general form of quadratic equation

3) 2x + 50 = 0., no x2 term, hence not a quadratic equation

4) x12 − 3 = 0. Exponent can’t be a fraction for a quadratic equation.

Hence, option 2 is correct answer


The given equation is f(x) = 2x2 + 5x + 3 = 0

f( − 2) = 2( − 2)2 + 5( − 2) + 3 = 8 − 10 + 3 = 1 ≠ 0

f( − 3) = 2( − 3)2 + 5( − 3) + 3 = 18 − 15 + 3 = 6 ≠ 0

f(4) = 2(42) + 5(4) + 3 = 32 + 20 + 3 = 55 ≠ 0

f( − 1) = 2( − 1)2 + 5( − 1) + 3 = 2 − 5 + 3 = 0

f( − 1) = 0, x = − 1 satisfies the given equation.


Put x = 3 in the given equation

√x2 − 4x + 3 + √x2 − 9 = √4x2 − 14x + 16, we get

√32 − 4(3) + 3 + √(32) − 9 = √4(32) − 14(3) + 16

√12 − 12 + √9 − 9 = √36 + 16 − 42

0 ≠ √10

L.H.S ≠ R.H.S Hence x = 3 is not a solution.


Given quadratic equation is x2 − 3x − 10 = 0

x2 − 5x + 2x − 10 = 0 x(x − 5) + 2(x − 5) = 0 (x − 5)(x + 2) = 0 x − 5 = 0 or x + 2 = 0 x = 5 or x = − 2.


Given quadratic equation is 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.


Given quadratic equation is x2 − 7x + 6 = 0 =x2 − 6x − x + 6 = 0 x(x − 6) − (x − 6) = 0 (x − 6)(x − 1) = 0 x − 6 = 0 or x − 1 = 0 x = 6 or x = 1.


Given quadratic equation is √2x2 + 7x + 5√2 = 0

=√2x2 + 5x + 2x + 5√2 = 0

⇒ x(√2x + 5) + √2(√2x + 5) = 0

⇒ (√2x + 5)(x + √2) = 0

⇒ √2x + 5 = 0, x + √2 = 0

⇒ x =− 5

√2 or x = − √2.


Given quadratic equation is 2x2 − x +18 = 0

=16x2 − 8x + 1

8 = 0

16x2 − 8x + 1 = 0 ⇒ (4x)2 − 2(4x)(1) + 12 = 0

[ ∵ a2 − 2ab + b2 = (a − b)2] ⇒ (4x − 1)2 = 0 ⇒ (4x − 1)(4x − 1) = 0

⇒ x =14 ,

14


Given quadratic equation is 4√3x2 + 5x − 2√3 = 0

4√3x2 + 8x − 3x − 2√3 = 0

4x(√3x + 2) − √3(√3x + 2) = 0

(√3x + 2)(4x − √3) = 0

4x − √3 = 0 or √3x + 2 = 0

x =√3

4 or x =− 2

√3

Hence the roots of given quadratic equation is √3

4 ,− 2

√3


Given quadratic equation is x2 − 4ax + 4a2 − b2 = 0

4a2 − b2 = (2a + b)(2a − b) and 4ax = (2a + b)x + (2a − b)x

⟹ x2 − [(2a + b)x + (2a − b)x] + (2a + b)(2a − b) = 0 Which is in the form x2 − (α + β) + αβ = 0 and α , β are the roots

⟹ 2a + b and 2a − b are the roots


Given quadratic equation is 100x2 − 20x + 1 = 0 100x2 − 10x − 10x + 1 = 0 10x(10x − 1) − 1(10x − 1) = 0 (10x − 1)(10x − 1) = 0

x =110 ,

110

Alternate method 100x2 − 20x + 1 = 0 (10x)2 − 2(10x)(1) + 12 = 0 (10x − 1)2 = 0

(10x − 1)(10x − 1) = 0

x =110 ,

110


Given quadratic equation is 3x2 − 2√6x + 2 = 0

3x2 − √6x − √6x + 2 = 0

√32x2 − √2√3x − √2√3x + √2√2

(√3x − √2)(√3x − √2) = 0

√3x − √2 = 0 or √3x − √2 = 0

x =√2

√3,

√2

√3


Given quadratic equation is

x −1x = 3

x2 − 1x = 3

x2 − 1 = 3x x2 − 3x − 1 = 0 Comparing with ax2 + bx + c = 0 a = 1, b = − 3, c = − 1

The roots of given quadratic equation are

x =− b±√b2 − 4ac

2a

=− ( − 3 ) ±√ ( − 3 )2 − 4 ( 1 ) ( − 1 )

2 ( 1 )

=3 ± √9 + 4

2

=3 ± √13

2

Hence the roots of given quadratic equation are

=3 + √13

2 ,3 − √13

2 .


Given quadratic equation is 1

x + 4 −1

x − 7 =1130

1x + 4 −

1x − 7 =

1130

x − 7 − x − 4(x + 4)(x − 7) =

1130

−11

x2 + 4x − 7x − 28=

1130

−11 × 30 = 11 × (x2 − 3x − 28)

0=x2 − 3x − 28 + 30

∴ x2 − 3x + 2 = 0

x2 − 2x − x + 2 = 0

x(x − 2) − 1(x − 2) = 0

(x − 2)(x − 1) = 0

x − 1 = 0, x − 2 = 0

x = 1, x = 2 Hence the roots of given quadratic equation are 1,2


Given equation is 1x −

1x − 3 =

43

⇒x − 3 − xx(x − 3) =

43

−3

x2 − 3x=

43

−9 = 4(x2 − 3x) ⇒ 4x2 − 12x + 9 = 0

4x2 − 6x − 6x + 9 = 0 ⇒ 2x(2x − 3) − 3(2x − 3) = 0

(2x − 3)(2x − 3) = 0

⇒ x =32 ,

32


Given equation is 1x −

1x− 2 = 3

x− 2 − xx ( x− 2 ) = 3

− 2

x2 − 2x=

31

On cross multiplication,we get −2 = 3(x2 − 2x) ⇒ 3x2 − 6x + 2 = 0

Comparing the above equation with ax2 + bx + c = 0,we get a = 3, b = − 6, c = 2 Discriminent D = b2 − 4ac = ( − 6)2 − 4 × 3 × 2 = 36 − 24 = 12 Which is >0,hence the roots of given quadratic equation are

x =− b± √D

2a =− ( − 6 ) ± √12

2.3 =6 ± 2√3

6

=2 ( 3 ± √3 )

6 =3 ± √3

3

Hence the roots of given quadratic equation are 3 + √3

3 ,3 − √3

3 .


Given equation is −3x2 + 2x − 8 = 0 Comparing the above equation with ax2 + bx + c = 0,we get a = − 3, b = 2, c = − 8 Discriminent D = b2 − 4ac = (2)2 − 4( − 3)( − 8) = 4 − 96 = − 92


Given equation is x +1x = 9

x2 + 1x = 9

⇒ x2 + 1 = 9x x2 − 9x + 1 = 0 Comparing the above equation with ax2 + bx + c = 0,we get a = 1, b = − 9, c = 1 Discriminent D = b2 − 4ac = ( − 9)2 − 4 × 1 × 1 = 81 − 4 = 77 Which is >0,hence the roots of given quadratic equation

exists and are x =− b± √D

2a = −− ( − 9 ) ± √77

2 × 1 =9 ± √77

2

Hence the roots of given quadratic equation are 9 + √77

2 ,9 − √77

2 .


Given equation is √3x2 − 2√2x − √12 = 0

Comparing the above equation with ax2 + bx + c = 0,we get a = √3, b = − 2√2, c = − √12

Discriminent D = b2 − 4ac = ( − 2√2)2 − 4(√3)( − √12) = 8 + 4√36

= 8 + 4 × 6 = 32

20: Option 4Solution Details Given equation is 4x2 − px + 2 = 0

Comparing the above equation with ax2 + bx + c = 0,we get a = 4, b = − p, c = 2 Discriminent D = b2 − 4ac = ( − p)2 − 4 × 4 × 2 = p2 − 32


Given equation is 2x2 − 5x − 7 = 0 Comparing the above equation with ax2 + bx + c = 0,we get a = 2, b = − 5, c = 7 Discriminent D = b2 − 4ac = ( − 5)2 − 4 × 2( − 7) = 25 + 56 = 81 > 0

Hence the roots of given quadratic equation are real and distinct.


Given equation is 2x2 − 4x + 3 = 0 Comparing the above equation with ax2 + bx + c = 0,we get a = 2, b = − 4, c = 3 Discriminent D = b2 − 4ac = ( − 4)2 − 4 × 2 × 3 = 16 − 24 = − 8 < 0

Hence,there is no real root for given equation.


Given equation is 4x2 − 4x + 1 = 0 Comparing the above equation with ax2 + bx + c = 0,we get a = 4, b = − 4, c = 1 Discriminent D = b2 − 4ac = ( − 4)2 − 4 × 4 × 1 = 16 − 16 = 0 Hence,the roots of given quadratic equation are real and equal.

24: Option 3Solution Details Given equation is 3x2 − 6x + 5 = 0

Comparing the above equation with ax2 + bx + c = 0,we get a = 3, b = − 6, c = 5 Discriminent D = b2 − 4ac = ( − 6)2 − 4 × 3 × 5 = 36 − 60 = − 24 < 0

Hence,the roots of given quadratic equation are not real.


Given equation is 3√3x2 + 10x + √3 = 0

Comparing the above equation with ax2 + bx + c = 0,we get a = 3√3, b = 10, c = √3

Discriminent D = b2 − 4ac = (10)2 − 4 × 3√3 × √3

= 100 − 36 = 64 > 0 Hence the roots of given quadratic equation are real and distinct.


Given equation is 2x2 + kx + 3 = 0 Comparing the above equation with ax2 + bx + c = 0,we get a = 2, b = k, c = 3 The roots of above equation are real and equal. i. e Discriminent D = b2 − 4ac = 0 Here k2 − 4.2.3 = 0 k2 − 24 = 0 ⇒ k = √24 = 2√6.


Given equation is 12x2 + 4kx + 3 = 0 Comparing the above equation with ax2 + bx + c = 0,we get a = 12, b = 4k, c = 3 The roots of above equation are real and equal. i. e Discriminent D = b2 − 4ac = 0 Here (4k)2 − 4 × 12 × 3 = 0

⇒ 16k2 − 144 = 0

k2 =14416 = 9

∴ k = ± 3


Given equation is x2 + 2kx + 9 = 0 Comparing the above equation with ax2 + bx + c = 0,we get a = 1, b = 2k, c = 9 The roots of above equation are real and equal. i. e Discriminent D = b2 − 4ac = 0 Here (2k)2 − 4 × 1 × 9 = 0 ⇒ 4k2 − 36 = 0

4k2 = 36 ∴ k = ± 3.


Given equation is px(3x − 4) + 4 = 0 ⇒ 3px2 − 4px + 4 = 0

Comparing the above equation with ax2 + bx + c = 0,we get a = 3p, b = − 4p, c = 4 The roots of above equation are real and equal. i. e Discriminent D = b2 − 4ac = 0 Here ( − 4p)2 − 4.3p.4 = 0 ⇒ 16p2 − 48p = 0

16p(p − 3) = 0 ∴ p = 0 or 3. ∵ for quadratic equation ax2 + 2bx + c = 0, a ≠ 0

So p ≠ 0 and p = 3 only.


Given equation is px(6x + 10) = − 25 ⇒ 6px2 + 10px + 25 = 0

Comparing the above equation with ax2 + bx + c = 0,we get a = 6p, b = 10p, c = 25 The roots of above equation are equal. i. e Discriminent D = b2 − 4ac = 0 Here (10p)2 − 4 × 6p × 25 = 0 ⇒ 100p2 − 600p = 0

100p(p − 6) = 0 ∴ p = 0 or P = 6. ∵ for quadratic equation ax2 + 2bx + c = 0, a ≠ 0

So p ≠ 0 and p = 6 only.


Given that -5 is a root of quadratic equation 2x2 + px − 15 = 0

⟹ 2( − 5)2 + p( − 5) − 15 = 0

⇒ 50 − 5p − 15 = 0

35 − 5p = 0

∴ p = 7

Substituting p = 7 , we get

7(x2 + x) + k = 0

⇒ 7x2 + 7x + k = 0.

The roots of above equation are equal Discriminant D = b2 − 4ac = 0

⇒ (7)2 − 4(7)(k) = 0

⇒ 49 − 28k = 0

⇒ k =4928 =

74 .


Given quadratic equation is (k + 4)x2 + (k + 1)x + 1 = 0 Comparing the above equation with ax2 + bx + c = 0,we get a = k + 4, b = k + 1, c = 1 The roots of above equation are equal. i. e Discriminant D=0 b2 − 4ac = 0 (k + 1)2 − 4(k + 4)(1) = 0 k2 + 2k + 1 − 4k − 16 = 0 k2 − 2k − 15 + 0 (k − 5)(k + 3) = 0 k − 5 = 0 or k + 3 = 0 k = 5 or k = − 3.


Let required numbers be x and 27 − x

[ ∵ sum of two numbers is x + (27 − x) = 27]

Given product of the numbers is 182 ⟹ x(27 − x) = 182

27x − x2 = 182

x2 − 27x + 182 = 0

x2 − 14x − 13x + 182 = 0

x(x − 14) − 13(x − 14) = 0

(x − 14)(x − 13) = 0

x − 14 = 0, x − 13 = 0

x = 14, x = 13.


Let the two consecutive positive integers be x and x + 1 Given the sum of squares=365 (x + 1)2 + (x)2 = 365 x2 + 2x + 1 + x2 = 365 2x2 + 2x − 364 = 0 x2 + x − 182 = 0 x2 + 14x − 13x − 182 = 0 x(x + 14) − 13(x + 14) = 0 (x + 14)(x − 13) = 0 x + 14 = 0, x − 13 = 0 x = − 14, x = 13. [ ∵ − 14 is not a positive integer] The required two consecutive positive integers are 13,14.


Let the two numbers be x and 48 − x

Given, Product of two numbers=432

⟹ x(48 − x) = 432

48x − x2 − 432 = 0

x2 − 48x + 432 = 0

x2 − 36x − 12x + 432 = 0

(x − 36)(x − 12) = 0

x − 36 = 0 or x − 12 = 0

x = 36

The other number is 48 − 36 = 12

∴ the required two numbers are 12 and 36.



sum of the reciprocals of two numbers=8

15

1x +

18 − x =

815

8 − x + xx(8 − x) =

815

8 × 15 = 8x(8 − x)

120 = 64x − 8x2

8x2 − 64x + 120 = 0

x2 − 8x + 15 = 0

x2 − 3x − 5x + 15 = 0 x(x − 3) − 5(x − 3) = 0 (x − 3)(x − 5) = 0

x − 3 = 0 or x − 5 = 0

x = 3, x = 5

If x = 3, the other is 8 − x

∴ the required two numbers are 3,5

Alternate proof

By hypothesis x + y = 8,1x +

1y =

815

⟹x + yxy =

815

8xy =

815

xy = 15 Required numbers are 3,5


Let the two consecutive odd positive integers be x and x + 2 Given sum of the squares two consecutive odd positive integers=290. ⟹ x2 + (x + 2)2 = 290 x2 + x2 + 2x + 4 = 290 2x2 + 2x + 4 = 290 on dividing by 2 ,we get

x2 + 2x + 2 = 145 x2 + 2x − 143 = 0 x2 + 13x − 11x − 143 = 0 (x − 11)(x + 13) = 0 ∴ x = 11

The required numbers are 11,13


Let the required two digit number be 10x + y Given,product of digits =10 ⇒ xy = 10--------(1)

When 27 is added from the number the digits intrchange their places. i. e 10x + y + 27 = 10y + x ⇒ 9x − 9y + 27 = 0 ⇒ x − y = − 3-------(2)

From equation (1), xy = 10 ⇒ x =10y

Put x =10y in equation(2),we get

10y − y = − 3

⇒10 − y2

y = − 3

⇒ 10 − y2 = − 3y 0 = y2 − 3y − 10 ∴ y2 − 3y − 10 = 0 ⇒ y2 − 5y + 2y − 10 = 0 ⇒ (y − 5)(y + 2) = 0 y − 5 = 0 or y + 2 = 0 ⇒ y = 5, y ≠ − 2

Put y = 5in equation-----(1),x =105 = 2

∴ the required number is 10x + y = 10(2) + 5 = 25.

VERIFICATION

For 25, product of digits = 2 × 5 and 25 + 27 = 52


Let the required two digit number be 10x + y

Given 10x + y = 4(x + y) ⇒ 10x − 4x + y − 4y = 0 ⇒ 6x − 3y = 0 ⇒ 2x − y = 0 ⇒ 2x = y------(1)

Given 10x + y = 2xy-------(2) Put y = 2x in equation (2),we get 10x + 2x = 2x(2x) ⇒ 12x = 4x2 ⇒ 4x2 − 12x = 0 ⇒ 4x(x − 3) = 0 ⇒ 4x = 0 or x − 3 = 0 x ≠ 0 so x = 3 Put x = 3 in equation (1).we get y = 2x = 2(3) = 6 ∴ the required number is 10x + y = 10(3) + 6 = 36


Let the two numbers be x, 9 − x

Sum of the reciprocals=1x +

19 − x =

12

⇒9 − x+ xx ( 9 − x ) =

12

⇒ 9 × 2 = x(9 − x) ⇒ 18 = 9x − x2 ⇒ x2 − 9x + 18 = 0 ⇒ x2 − 6x − 3x + 18 = 0 ⇒ (x − 6)(x − 3) = 0 ⇒ x − 6 = 0 or x = 3

If x = 3 the other number is 9 − 3 = 6. ∴ the required numbers or 3,6


Let the required number be x then its reciprocal is 1x .

Given x +1x = 2

120

⇒x2 + 1x =

4120

⇒ 20x2 + 20 = 41x ⇒ 20x2 − 41x + 20 = 0 ⇒ 20x2 − 25x − 16x + 20 = 0 ⇒ 5x(4x − 5) − 4(4x − 5) = 0 ⇒ (4x − 5)(5x − 4) = 0 ⇒ 4x − 5 = 0 or 5x − 4 = 0

⇒ x =54 ,

45

54 ,

45 are reciprocals.


Let the required number be x then its reciprocal is 1x .

Given x +1x =

507

⇒x2 + 1x =

507

⇒ 7x2 + 7 = 50x ⇒ 7x2 − 50x + 7 = 0 ⇒ 7x2 − 49x − x + 7 = 0 ⇒ 7x(x − 7) − 1(x − 7) = 0 ⇒ (x − 7)(7x − 1) = 0 ⇒ x − 7 = 0 or 7x − 1 = 0 ⇒ x = 7, 17


Let required numbers be x and 17 − x [ ∵ sum of two numbers is 17. x + (17 − x) = 17 x(17 − x) = 72 17x − x2 = 72 x2 − 17x + 72 = 0 x2 − 9x − 8x + 72 = 0 x(x − 9) − 8(x − 9) = 0 (x − 9)(x − 8) = 0 x − 9 = 0, x − 8 = 0 x = 9, x = 8.


Given quadratic equation is x2 − 5x + q = 0 Put x = 3 in the above equation,we get 32 − 5(3) + q = 0 ⇒ 9 − 15 + q = 0 ⇒ − 6 + q = 0 ∴ q = 6.


Given equation is x2 − x + 2 = 0 Comparing the above equation with ax2 + bx + c = 0,we get a = 1, b = 1, c = 2 Discriminent D = b2 − 4ac = ( − 1)2 − 4 × 1 × 2

= 1 − 8 = − 7 < 0 Hence there are no real roots for the given quadratic equation.


Given equation is 4x2 − 12x + 9 = 0 Comparing the above equation with ax2 + bx + c = 0,we get a = 4, b = − 12, c = 9 Discriminent D = b2 − 4ac = ( − 12)2 − 4 × 4 × 9 = 144 − 144 = 0 Hence,the roots of given quadratic equation are real and equal.


Let present ages of Varun and Swati be x and y years. Seven years ago their ages were x − 7, y − 7 Given x − 7 = 5(y − 7)2 ⇒ x − 7 = 5(y2 − 14 + 49)---------(1)

Three years hence their ages will be x + 3, y + 3

Given y + 3 =25 (x + 3)

⇒ 5y + 15 = 2x + 6 ⇒ 5y + 9 = 2x

⇒ x =5y+ 9

2

Put x =5y+ 9

2 in equation------(1),we get 5y+ 9

2 − 7 = 5(y2 − 14y + 49) 5y+ 9 − 14

2 = 5((y2 − 14y + 49)

⇒5y− 5

2 = 5((y2 − 14y + 49)

⇒y− 1

2 = y2 − 14y + 49

⇒ y − 1 = 2y2 − 28y + 98 ⇒ 2y2 − 28y + 98 − y + 1 = 0 ⇒ 2y2 − 29y + 99 = 0 ⇒ 2y2 − 18y − 11y + 99 = 0 ⇒ 2y(y − 9) − 11(y − 9) = 0 ⇒ (2y − 11)(y − 9) = 0 ⇒ 2y − 11 = 0, y − 9 = 0

⇒ y =112 , y = 9

[ ∵ age can’t be a fraction y is ≠112 ]

∴ y = 9, x =5y+ 9

2 =5.9 + 9

2 =542 = 27

x = 27, y = 9

48: Option 1

Solution Details


The sum of reciprocals=310

i. e 1x +

115 − x =

310

⇒15 − x+ xx ( 15 − x ) =

310

⇒ 150 = 3x(15 − x) 50 = x(15 − x) ⇒ 50 = 15x − x2 ⇒ x2 − 15x + 50 = 0 ⇒ x2 − 10x − 5x + 50 = 0 ⇒ x(x − 10) − 5(x − 10) = 0 ⇒ (x − 10)(x − 5) = 0 ⇒ x − 10 = 0, x − 5 = 0 ⇒ x = 10, 5


Let the two numbers be x and x + 2

The difference of reciprocals=14

i. e1x −

1x+ 2 =

14

⇒x+ 2 − xx ( x+ 2 ) =

14

⇒ 8 = x(x + 2) 8 = x2 + 2x ⇒ x2 + 2x − 8 = 0 ⇒ x2 + 4x − 2x − 8 = 0 ⇒ x(x + 4) − 2(x + 4) = 0 ⇒ (x + 4)(x − 2) = 0 ⇒ x + 4 = 0 or x − 2 = 0

[ ∵ − 4 is not a natural number x ≠ − 4] Hence the required numbers are 2,4.


Let the two consecutive positive integers be x and x + 1 Given sum of the squares of the two numbers is 481. i. e x2 + (x + 1)2 = 481 ⇒ x2 + x2 + 2x + 1 − 481 = 0 ⇒ 2x2 + 2x − 480 = 0 ⇒ x2 + x − 240 = 0 ⇒ x2 + 16x − 15x − 240 = 0 ⇒ x(x + 16) − 15(x + 16) = 0 ⇒ (x + 16)(x − 15) = 0 ⇒ x + 16 = 0 or x − 15 = 0 x = − 16, 15

[ ∵ − 16 is not a positive integers,x ≠ − 16 Hence the required two consecutive positive integers are 15,16.


Let the two consecutive positive even numbers be 2x, 2x + 2 Given the sum of the squares of two consecutive positive even numbers is 244. i. e (2x)2 + (2x + 2)2 = 244 ⇒ 4x2 + 4x2 + 8x + 4 = 244. ⇒ 8x2 + 8x + 4 − 244 = 0 ⇒ 8x2 + 8x − 240 = 0 ⇒ x2 + x − 30 = 0 ⇒ x2 + 6x − 5x − 30 = 0 ⇒ x(x + 6) − 5(x + 6) = 0 ⇒ (x + 6)(x − 5) = 0 ⇒ x + 6 = 0, x − 5 = 0 ⇒ x = − 6, 5

[ ∵ − 6 is not positive x ≠ − 6] So, x = 5 and the required two numbers are 2(5), 2(5) + 2 = 10, 12 ∴ the smallest number is 10.


Let the ages of Raju’s father and Raju be x and (x − 30) years. Given difference of the squares their ages=1560. i. e x2 − (x − 30)2 = 1560 ⇒ x2 − (x2 − 60x + 900) = 1560 ⇒ x2 − x2 + 60x − 900 = 1560 ⇒ 60x = 1560 + 900 = 2460

⇒ x =246060 = 41

∴ father’s age =x = 41 Raju’s age=x − 30 = 41 − 30 = 11.


Given , x = √2 is a solution of the equation kx2 + √2x − 4 = 0

Put x = √2 in kx2 + √2x − 4 = 0

⇒ k(√2)2 + √2(√2) − 4 = 0

2k + 2 − 4 = 0 ⇒ 2k − 2 = 0 ⇒ k = 1.


Given equation is x − 4x − 5 +

x − 6x − 7 =

103

⇒(x − 4)(x − 7) + (x − 6)(x − 5)

(x − 5)(x − 7) =103

⇒x2 − 11x + 28 + x2 − 11x + 30

x2 − 12x + 35=

103

⇒2x2 − 22x + 58

x2 − 12x + 35=

103

On cross multiplication, 6x2 − 66x + 174 = 10x2 − 120x + 350 ⇒ 0 = 4x2 − 54x + 176 = 0

4x2 − 54x + 176 = 0 ⇒ 2x2 − 27x + 88 = 0

2x2 − 16x − 11x + 88 = 0 2x(x − 8) − 11(x − 8) = 0 (x − 8)(2x − 11) = 0 ⇒ x − 8 = 0 or 2x − 11 = 0

∴ x = 8,112


The given quadratic equation is 4x − 32x + 1 − 10(

2x + 14x − 3 ) = 3

(4x − 3)2 − 10(2x + 1)2

(2x + 1)(4x − 3) =31

16x2 + 9 − 24x − 40x2 − 40x − 10

8x2 − 2x − 3=

31

−24x2 − 64x − 1 = 24x2 − 6x − 9 −24x2 − 24x2 − 64x + 6x − 1 + 9 = 0 −48x2 − 58x + 8 = 0 48x2 + 58x − 8 = 0 24x2 + 29x − 4 = 0 8x(3x + 4) − 1(3x + 4) = 0 (3x + 4)(8x − 1) = 0 3x + 4 = 0 or 8x − 1 = 0

x = −43 , x =

18


Given equation is 1

x − 1 −1

x + 5 =67

⇒x + 5 − x + 1(x − 1)(x + 5) =

67

⇒ 6 × 7 = 6(x − 1)(x + 5) 7 = x2 − x + 5x − 5 x2 + 4x − 12 = 0

⇒ x2 + 6x − 2x − 12 = 0 x(x + 6) − 2(x + 6) = 0 (x − 2)(x + 6) = 0 ⇒ x − 2 = 0 or x + 6 = 0 x = 2, − 6.



x − 3x − 4 =

103

⇒(x − 1)(x − 4) + (x − 3)(x − 2)

(x − 2)((x − 4) =103

⇒(x2 − 5x + 4) + (x2 − 5x + 6)

(x2 − 6x + 8)=

103

⇒ 6x2 − 30x + 30 = 10x2 − 60x + 80 6x2 − 10x2 − 30x + 60x + 30 − 80 = 0 ⇒ − 4x2 + 30x − 50 = 0 ⇒ − 2(2x2 − 15x + 25) = 0 ⇒ (2x2 − 15x + 25) = 0

2x2 − 10x − 5x + 25 = 0 2x(x − 5) − 5(x − 5) = 0 (x − 5)(2x − 5) = 0 ⇒ x − 5 = 0 or 2x − 5 = 0

∴ x = 5,52



x − 4x − 5 =

103

⇒(x − 2)(x − 5) + (x − 4)(x − 3)

(x − 3)(x − 5) =103

⇒(x2 − 7x + 10) + (x2 − 7x + 12)

(x2 − 8x + 15)=

103

3(2x2 − 14x + 22) = 10(x2 − 8x + 15) 6x2 − 42x + 66 = 10x2 − 80x + 150

6x2 − 10x2 − 42x + 80x + 66 − 150 = 0 −4x2 + 38x − 84 = 0 −2(2x2 − 19x + 42) = 0 ⇒ 2x2 − 19x + 42 = 0

2x2 − 12x − 7x + 42 = 0 2x(x − 6) − 7(x − 6) = 0 (x − 6)(2x − 7) = 0 ⇒ x − 6 = 0 or 2x − 7 = 0

∴ x = 6,72 .


GIven quadratic equation is x2 − 2x(1 + 3k) + 7(3 + 2k) = 0 Compare the above equation with ax2 + bx + c = 0, we get a = 1, b = − 2(1 + 3k), c = 7(3 + 2k) Given quadratic equation has equal roots. i. e Discriminent ‘D’=0 ⇒ b2 − 4ac = 0 ⇒ ( − 2(1 + 3k))2 − 4(1) × 7(3 + 2k) = 0

4(1 + 9k2 + 6k) − 28(3 + 2k) = 0 4 + 36k2 + 24k − 84 − 56k = 0 36k2 − 32k − 80 = 0 ⇒ 4(9k2 − 8k − 20) = 0 ⇒ 9k2 − 8k − 20 = 0 ⇒ 9k2 − 18k + 10k − 20 = 0

9k(k − 2) + 10(k − 2) = 0 (k − 2)(9k + 10) = 0 k − 2 = 0 or 9k + 10 = 0

∴ k = 2,− 10

9


Given quadratic equation is k2x2 − 2(k − 1)x + 4 = 0 Compare the above equation with ax2 + bx + c = 0, we get a = k2, b = − 2(k − 1), c = 4 The given equation has real and equal. i. e, Discriminent, D = 0 b2 − 4ac = 0 ( − 2(k − 1))2 − 4(k2)(4) = 0 4(k − 1)2 − 4(4k2) = 0 4[(k − 1)2 − 4k2] = 0

k2 − 2k + 1 − 4k2 =04 = 0

−3k2 − 2k + 1 = 0 3k2 + 2k − 1 = 0 3k2 + 3k − k − 1 = 0 3k(k + 1) − 1(k + 1) = 0 (k + 1)(3k − 1) = 0 k + 1 = 0 or 3k − 1 = 0

k = − 1, k =13


Given quadratic equation is 2mx2 − 2(1 + 2m)x + (3 + 2m) = 0 Compare the above equation with ax2 + bx + c = 0, we get a = 2m, b = − 2(1 + 2m), c = 3 + 2m Given equation has real and distinct roots.

i. e Discriminanat D>0 b2 − 4ac > 0 ( − 2(1 + 2m))2 − 4(2m)(3 + 2m) > 0 4(1 + 2m)2 − 4(6m + 4m2) > 0 4(1 + 4m + 4m2) − 4(6m + 4m2) > 0 4 + 16m + 16m2 − 24m − 16m2 > 0 4 − 8m > 0 4(1 − 2m) > 0 1 − 2m > 0or 2m − 1 < 0

m <12

For m <12 the given equation has real and distinct roots.


Let the age of his son one year ago be x and father’s age=8x Present age of son=x + 1 present age of father=8x + 1 8x + 1 = (x + 1)2 8x + 1 = x2 + 2x + 1 0 = x2 − 2x − 8x x2 − 6x = 0 x(x − 6) = 0 x = 0, x − 6 Hence the present age of son is x + 1 = 6 + 1 = 7years

Present age of man 8x + 1 ⇒ 8(6) + 1 = 49 years.


Let the speed of the train be x k.m.p.h. Distance traveled d = 360km

Time t1 =distancetime =

360x ------(1)

If the speed had been increased by 5k.m.p.h, speed=x + 5

Time t2 =distancespeed =

360x+ 5 -------(2)

Given that if the speeds increased by 5 k.m.p.h, it would have taken 1 hour less time forthe same journey. i. e t1 − t2 = 1

From equation (1) and (2) t1 − t2 =360x −

360x+ 5 = 1

⇒ 360(x+ 5 − xx ( x+ 5 ) ) = 1

⇒1800

x ( x+ 5 ) = 1

⇒ 1800 = x(x + 5) ⇒ x2 + 5x − 1800 = 0 ⇒ x2 + 45x − 40x − 1800 = 0 x(x + 45) − 40(x + 45) = 0

(x + 45)(x − 40) = 0 ⇒ x + 45 = 0, x − 40 = 0 x = − 45, 40 [ ∵ speed can’t be negative x ≠ − 45 ⇒ x = 40 ∴ The speed of the train is 40k.m.p.h.


Let the actual speed of the plane be x k.m.p.h Difference d = 1250 km

Actual time(ta) =ds

ta =1250x -------(1)

Now the speed is increased by 250k.m.p.h v = (x + 250)k.m.p.h d = 1250

New time tn =ds

⇒ tn =1250x + 250 -------(2)

ta − tn = 50minutes

⇒1250x −

1250x + 250 = 50 ×

160 hours

1250(1x −

1x + 250 ) =

56

1250(x + 250 − xx(x + 250) ) =

56

⇒1250(250)

x2 + 250x=

56

65 × 1250(250) = x2 + 250x

⇒ 6 × 62500 = x2 + 250x

⇒ 375000 = x2 + 250x

⇒ x2 + 250x − 375000 = 0

⇒ x2 + 750x − 500x − 375000 = 0

⇒ x(x + 750) − 500(x + 750) = 0 (x + 750)(x − 500) = 0 x + 750 = 0 x = − 750 [ ∵ speed can’t be negative] (or)

x − 500 = 0 x = 500 Actual speed of plane =500 k.m.p.h.


Let the speed of the stream=x k.m.p.h Speed of the boat upstream=(8 − x) k.m.p.h Speed of the boat downstream=(8 + x) k.m.p.h

Time taken in up stream(tu) =distance travelled in up stream

speed

tu =15

8 − x ------(1)

Time taken in down stream =distance travellled in down stream

speed

=22

8 + x

Total time taken =5 h tu + td = 5

158 − x +

228 + x = 5

15(8 + x) + 22(8 − x)(8 − x)(8 + x) = 5

120 + 15x + 176 − 22x = 5(64 − x2) −7x + 296 = 320 − 5x2 5x2 − 7x − 320 + 296 = 0 5x2 − 7x − 24 = 0 ⇒ 5x2 + 8x − 15x − 24 = 0 x(5x + 8) − 3(5x + 8) = 0 (x − 3)(5x + 8) = 0 5x + 8 = 0

x = −85 [ ∵ speed can’t be negative] x ≠ −

85

∴ speed of stream x − 3 = 0 x = 3 k.m.p.h.

66: Option 2Solution Details Let the shorter side be x m

Longer side (l) = x + 30 Diagonal (d) = x + 60

We know that hypotenuse²=base²+altitude² d2 = b2 + l2 (x + 60)2 = x2 + (x + 30)2 x2 + 120x + 3600 = x2 + x2 + 60x + 900 x2 + 120x + 3600 − x2 − x2 − 60x − 900 = 0

−x2 + 60x + 2700 = 0 x2 − 60x − 2700 = 0 x2 − 90x + 30x − 2700 x(x − 90) + 30(x − 90) = 0 (x − 90)(x + 30) = 0 x − 90 = 0 or x + 30 = 0 x = 90, x = − 30 [ ∵ length can’t be negative] x ≠ − 30 Length(l)=x + 30 = 90 + 30 = 120 Breadth(b)=x = 90 Dimension of field are 120 × 90m


Let the breadth of plot be x m Then length=2x + 1 m

Area ‘A’=528 Given in the problem l × x = 528 (2x + 1)x = 528 2x2 + x − 528 = 0 ⇒ 2x2 + 33x − 32x − 528 = 0 x(2x + 33) − 16(2x + 33) = 0 ⇒ (x − 16)(2x + 33) = 0 x − 16 = 0 or 2x + 33 = 0

x = 16, x = −332

[ ∵ breadth can’t be negative]

x ≠ −332

length=2x + 1 = 2(16) + 1 = 33 breadth =x = 16.


Let the sides be of two squares x and y Given in the problem sum of the area’s of two squares is 640m². ∴ x2 + y2 = 640----------(1)

Given in the problem difference in perimeter is 4x − 4y = 64 Divide by ‘4’ on both sides x − y = 16 x − 16 = y--------(2) Put y = x − 16 in-------(1) x2 + (x − 16)2 = 640 ⇒ x2 + x2 − 32x + 256 − 640 = 0

2x2 − 32x − 384 = 0 x2 − 16x − 192 = 0 ⇒ x2 − 24x + 8x − 192 = 0 x(x − 24) + 8(x − 24) = 0 ⇒ (x − 24)(x + 8) = 0

x − 24 = 0 or x + 8 = 0 x = 24, x = − 8 [ ∵ length can’t be negative.] x ≠ − 8 So, x = 24, then y = x − 16 = 24 − 16 = 8 The length of sides of two squares are 24 and 8m.


Let the breadth of rectangle be xcm Given, area ‘A’=240 ⇒ length × breadth = 240 x(x + 8) = 240 x2 + 8x − 240 = 0 ⇒ x2 + 20x − 12x − 240 = 0 x(x + 20) − 12(x + 20) = 0 ⇒ (x − 12)(x + 20) = 0 x − 12 = 0, x + 20 = 0 [ ∵ length can’t be negative] x ≠ − 20 ∴ x = 12 length = x + 8 = 12 + 8 = 20cm breadth = x = 12cm


1) (x + 1)2 = 2(x − 3) ⇒ x2 + 2x + 1 = 2x − 6 ⇒ x2 + 2x − 2x + 1 + 6 = 0 ⇒ x2 + 7 = 0 x2 + 0(x) + 7 = 0 Which is in the form ax2 + bx + c = 0 and a ≠ 0.Hence the given equationis quadratic equation.

2) (x − 2)(x + 4) = (x − 7)(x + 1) on simplification ⇒ x(x + 4) − 2(x + 4) = x(x + 1) − 7(x + 1) ⇒ x2 + 4x − 2x − 8 = x2 + x − 7x − 7 ⇒ x2 + 2x + 6x − 8 + 7 − x2 = 0 ⇒ 8x − 1 = 0 which is not in the form ofax2 + bx + c = 0and a ≠ 0

hence the given equation is not a quadratic equation.

3) (x + 2)3 = 4x(2x2 − 7)

⇒ x3 + 6x2 + 12x + 8 = 8x3 − 28x ∵ (a + b)3 = a3 + 3a2b + 3ab2 + b3

⇒ x3 − 8x3 + x2 + 12x + 28x + 8 = 0 ⇒ − 7x3 + 6x2 + 40x + 8 = 0

which is not in the form ax2 + bx + c = 0 hence the given equation is not a quadratic equation

[ ]


1) x3 − 4x2 − x + 1 = (x − 2)3

⇒ x3 − 4x2 − x + 1 = x3 − 6x2 + 12x − 8 ∵ (a − b)3 = a3 − 3a2b + 3ab2 − b3

⇒ x3 − 4x2 − x + 1 − x3 + 6x2 − 12x + 8 = 0 ⇒ 2x2 − 13x + 9 = 0, which is in the form ax2 + bx + c = 0.

Hence the given equation is a quadratic equation.

2) x +1x = 3, x ≠ 0

Multiplying by x on both sides, we get ⇒ x2 + 1 = 3x ⇒ x2 − 3x + 1 = 0, Which is in the form ax2 + bx + c = 0 and a ≠ 0.Hence the given

equation is quadratic equation.

3) 4x − 3 =

52x + 3

⇒4x −

31 =

52x + 3

⇒4 − 3xx =

52x + 3

on cross multiplication, we get(4 − 3x)(2x + 3) = 5x ⇒ 4(2x + 3) − 3x(2x + 3) = 5x ⇒ 8x + 12 − 6x2 − 9x = 5x ⇒ − 6x2 − x + 12 − 5x = 0 ⇒ − 6x2 − 6x + 12 = 0 ⇒ − 6(x2 + x − 2) = 0

⇒ x2 + x − 2 =0

− 6 = 0, Which is in the form ax2 + bx + c = 0 and a ≠ 0.Hence the given


[ ]


1) x− 1x+ 2 +

x+ 2x− 1 =

79

On simplification

⇒( x− 1 )2 + ( x+ 2 )2

( x+ 2 ) ( x− 1 ) =79

⇒x2 − 2x+ 1 + x2 + 4x+ 4

x2 + 2x− x− 2=

79

⇒2x2 + 2x+ 5

x2 + x− 2=

79

on cross multiplication, we get ⇒ 9(2x2 + 2x + 5) = 7(x2 + x − 2) ⇒ 18x2 + 18x + 45 = 7x2 + 7x − 14 ⇒ 18x2 − 7x2 + 18x − 7x + 45 + 14 = 0 ⇒ 11x2 + 11x + 59 = 0, Which is in the form ax2 + bx + c = 0 and a ≠ 0.Hence the given


2) x+ 1x− 1 +

x− 1x+ 1 =

247

On simplification

⇒( x+ 1 )2 + ( x− 1 )2

( x− 1 ) ( x+ 1 ) =247

⇒( x+ 1 )2 + ( x− 1 )2

x2 − 1=

247

⇒x2 + 2x+ 1 + x2 − 2x+ 1

x2 − 1=

247

⇒( 2x2 + 2 )

x2 − 1=

247

⇒2 ( x2 + 1 )

x2 − 1=

247

⇒ 7(x2 + 1) = 12(x2 − 1) ⇒ 7x2 + 7 = 12x2 − 12 ⇒ 12x2 − 7x2 − 12 − 7 = 0 ⟹ 5x2 − 19 = 0, which is in the form of ax2 + bx + c = 0 and a ≠ 0 Hence, the given equation is a quadratic equation.


Given quadratic equation is x2 − 2ax + (a2 − b2) = 0 Comparing the above equation with ax2 + bx + c = 0,we get a = 1, b = − 2a, c = a2 − b2 Discriminent D = b2 − 4ac = ( − 2a)2 − 4(1)(a2 − b2) = 4a2 − 4a2 + 4b2 = 4b2

Which is >0,hence the roots of given quadratic equation are

x =− b± √D

2a =− ( − 2a ) ±√4b2

2.1 =2a± 2b

2

⇒ x =2a+ 2b

2 or x =2a− 2b

2

⇒ x = a + b or x = a − b Hence the roots of given quadratic equation are a + b, a − b.


Given quadratic equation is 2x2 + ax − a2 = 0 Comparing the above equation with ax2 + bx + c = 0,we get a = 2, b = a, c = − a2 Discriminent D = b2 − 4ac = a2 − 4(2)( − a2) = a2 + 8a2 = 9a2

Which is >0,hence the roots of given quadratic equation


2a =− a±√9a2

2.2 =− a± 3a

4

⇒ x =− a+ 3a

4 or x =− a− 3a

4

⇒ x =a2 or x = − a

Hence the roots of given quadratic equation are a2 , − a.


Given quadratic equation is 3x2 − 5x + 2 = 0. Comparing the above equation with ax2 + bx + c = 0,we get a = 3, b = − 5, c = 2. Discriminent D = b2 − 4ac = ( − 5)2 − 4.3.2 = 25 − 24 = 1 Which is >0,hence the roots of given quadratic equation


2a =− ( − 5 ) ± √1

2.3

⇒ x =5 + 1

6 or x =5 − 1

6

x = 1 or x =23

Hence the roots of given quadratic equation are 1,23 .


Given quadratic equation is 4x2 + 3x + 5 = 0.

(2x)2 + 2(2x)(34 ) = − 5

Add (34 )2

(2x)2 + 2(2x)(34 ) + (

34 )2 = − 5 + (

34 )2

(2x +34 )2 = − 5 +

916 =

− 80 + 916 =

− 7116 < 0.

Which is not posssible since (2x +34 )2can’t be negative for any real value of x.

So,there is no solution for the given quadratic equation.


Given quadratic equation is 4x2 + 4√3x + 3 = 0

(2x)^2+2(2x)(\sqrt{3})+(\sqrt{3})^2=0 \Rightarrow (2x+\sqrt{3})^2=0 (2x+\sqrt{3})=0 x=\frac{-\sqrt{3}}{2} Hence the roots of given quadratic equation are \frac{-\sqrt{3}}{2},\frac{-\sqrt{3}}{2}.


Given quadratic equation is 2x^2-7x+3=0 Multiplying by 2 on both sides,we get \Rightarrow 4x^2-14x+6=0 \Rightarrow 4x^2-2(2x)(\frac{7}{2})+6=0 Adding and substracting (\frac{7}{2})^2 \Rightarrow (2x)^2-2(2x)(\frac{7}{2})+(\frac{7}{2})^2-(\frac{7}{2})^2+6=0 \Rightarrow (2x-\frac{7}{2})^2-\frac{49}{4}+6=0 \left[\because a^2-2ab+b^2=(a-b)^2\right]

\Rightarrow (2x-\frac{7}{2})^2=\frac{49}{4}-6 \Rightarrow (2x-\frac{7}{2})^2=\frac{49-24}{4}=\frac{25}{4} \Rightarrow 2x-\frac{7}{2}=\pm\frac{5}{2} \Rightarrow 2x=\frac{7}{2}+\frac{5}{2} or 2x= \frac{7}{2}-\frac{5}{2} \Rightarrow 2x=\frac{12}{2} or 2x=\frac{2}{2} \Rightarrow x=3 or x=\frac{1}{2}


Given that sum of the areas of two squares is \ 656\ \ cm^2

Area of square = (side)²

i.e. x^2+(x+4)^2=656 \implies x^2+x^2+8x+16-656=0 \implies 2x^2+8x-640=0 \implies x^2+4x-320=0 \implies x^2+20x-16x-320=0 \implies x(x+20)-16(x+20)=0 \implies (x+20)(x-16)=0 \implies x+20=0,x-16=0 \implies x=-20,x=16 \because\ length can’t be negative \ \implies x \neq -20 \therefore \ x=16 \therefore \ The sides of the two squares are \ x=16 , x+4=16+4=20


Let the altitude (h) of the triangle be \ x\ cm. Then , base (b) = 10+x Given that area ‘A’ = 600 cm² \implies \displaystyle\frac{1}{2}\times base\times height =600 cm^2 \implies \displaystyle\frac{1}{2} \times (10+x) \times x=600 \implies (10+x)x=1200 \implies x^2+10x-1200=0 \implies x^2+40x-30x-1200=0 \implies x(x+40)-30(x+40)=0 \implies (x+40)(x-30)=0 \implies x+40=0 \ or \ x-30=0 \implies x=-40,x=30 \because length can’t be negative \ \implies x\neq -40 \therefore x=30 Then , base (b) = x+10=30+10=40\ cm Altitude (h) = x=30\ cm


Given that \ y=1\ is the common root for both the equations \ ay^2+ay+3=0\ and \

y^2+y+b=0 \therefore \ y=1 \ satisfies both the equations . Substituting \ y = 1\ in both the equations , we have a(1)^2+a(1)+3=0 \implies 2a+3=0 \implies a=\displaystyle -\frac{3}{2} And \ 1^2+1+b=0 \implies b=-2 Now , \ a \times b =\displaystyle-\frac{3}{2} \times -2 =+\frac{6}{2}=3


Given x^2-3x+1=0 \implies x^2+1=3x \implies \displaystyle\frac{x^2+1}{x}=3 \implies \displaystyle\frac{x^2}{x}+\frac{1}{x}=3 \implies x+\displaystyle\frac{1}{x}=3


Given equation is \ 81x^2-1=0 \implies 81x^2=1 \implies x^2=\displaystyle\frac{1}{81} \implies x=\sqrt{\displaystyle\frac{1}{81}}=\pm \displaystyle\frac{1}{9} \displaystyle x = \frac{1}{9}


Given quadratic equation is \ x^2+kx-\displaystyle\frac{5}{4}=0 Substituting \ x=\displaystyle\frac{1}{2}\ in the above equation , we have \displaystyle \left(\frac{1}{2}\right)^2+k\left(\frac{1}{2}\right)-\frac{5}{4}=0

\implies \displaystyle\frac{1}{4}+\frac{k}{2}-\frac{5}{4}=0

\implies \displaystyle\frac{1+2k-5}{4}=0

\implies \displaystyle\frac{2k-4}{4}=0

\implies 2k-4=0 \implies 2k=4 \implies k=2 \therefore \displaystyle\frac{1}{2} will be a root of given equation when \ k=2 .


Given that the roots of \ 6kx^2+10kx+25=0\ are not real .

i.e., Discriminant \ D < 0 \implies b^2-4ac<0 \implies (10k)^2-4(6k)(25)<0 100k^2-600k <0 100k(k-6)<0 100k<0 \ or \ k-6<0 \implies k<0,k<6 we consider k<6


Given that \ kx^2+6x+9=0\ has no real roots .

i.e., Discriminant D < 0 \implies b^2-4ac< 0 \implies 6^2-4(k)(9) < 0 \implies 36-36k < 0 \implies 36(1-k)<0 1-k < 0 \implies 1< k \therefore k>1


Let us assume that ‘B’ takes \ x \ days to finish the work . \therefore \ ‘B’ can do \ \left(\displaystyle\frac{1}{x}\right)^{th} \ part of work in one day . Given , A takes \ (x-6) days to finish the work . \therefore \ A can do \ \left(\displaystyle\frac{1}{x-6}\right)^{th} \ part of work in one day . Both together they can finish the work in 4 days.

i.e., \ \left(\displaystyle\frac{1}{x}\right)4+\left(\frac{1}{x-6}\right)4=1

\implies 4\left(\displaystyle\frac{x-6+x}{x(x-6)}\right)=1

\implies 4(2x-6)=x(x-6) 8x-24=x^2-6x x^2-6x-8x+24=0 \implies x^2-14x+24=0 \implies x^2-12x-2x+24=0 \implies x(x-12)-2(x-12)=0 \implies (x-2)(x-12)=0 \implies (x-2)=0 \ or\ (x-12)=0 \implies x=2 \ , \ x=12 But , \ x \ cannot be less than 6 . So , \ x=12 Hence , B alone can finish the work in 12 days .


Let the two pipes take the time \ x\ minutes and \ x+3\ minutes respectively Each pipe can fill \ \displaystyle \left(\frac{1}{x}\right)^{th}\ and \ \displaystyle \left(\frac{1}{x+3}\right)^{th}\ part of a tank in one minute . If both are working together for \ 3\frac{1}{13}\ minutes to fill the tank , then we have \implies \left(\displaystyle\frac{1}{x}+\frac{1}{x+3}\right) \times 3\frac{1}{13}=1

\implies \displaystyle\frac{x+3+x}{x(x+3)} \times \frac{40}{13}=1

\implies (2x+3)40=13x(x+3) \implies 80x+120=13x^2+39x \implies 0=13x^2+39x-80x-120 \implies 13x^2-41x-120=0 \implies 13x^2-65x+24x-120=0 \implies 13x(x-5)+24(x-5)=0 \implies (x-5)(13x+24)=0 \implies x-5=0 \ or\ 13x+24=0 \implies x=5 And \ 13x=-24 ; x=\displaystyle\frac{-24}{13} \because \ time can’t be negative \ \implies x \neq \displaystyle\frac{-24}{13} \therefore Time taken by the two pipes is x = 5 \ minutes and \ x+3=5+3=8 \ minutes .


Let the time taken by the two pipes to fill the tank separately be \ x \ minutes and \ x+5\minutes respectively . They can fill \ \displaystyle \left(\frac{1}{x}\right)^{th}\ and \ \displaystyle \left(\frac{1}{x+5}\right)^{th} \ part of total tank in one minute . According to the given problem if both together are working for \ \displaystyle 11\frac{1}{9}\ minutes they can fill the tank .

i.e., \ \left(\displaystyle\frac{1}{x}+\frac{1}{x+5}\right)11\frac{1}{9}=1

\implies \left(\displaystyle\frac{x+5+x}{x(x+5)}\right)\displaystyle\frac{100}{9}=1

\implies (2x+5)100=9x(x+5) \implies 200x+500=9x^2+45x \implies 0=9x^2+45x-200x-500 \implies 9x^2-155x-500=0 \implies 9x^2-180x+25x-500=0 \implies 9x(x-20)+25(x-20)=0 \implies (x-20)(9x+25)=0 \implies x-20=0 \ or \ 9x+25=0 \implies x=20 \ , \ x=-\displaystyle\frac{25}{9}

\because \ time can’t be negative \ \implies x\neq -\displaystyle\frac{25}{9}

\therefore x=20 \ minutes \therefore \ Time taken by two taps to fill the tank separately is \ x=20 \ minutes and \x+5=20+5=25 \ minutes .


Let the number of books be \ x \ . Number of books \ \times \ cost of each book = Rs.80 \implies Cost of each book = \displaystyle\frac{80}{x} \ -----------(1)

If number of books increased by ‘4’ then cost of each book is decreased by Re.1.

\implies \left(\displaystyle\frac{80}{x}-1\right)(x+4)=80

\implies \left(\displaystyle\frac{80-x}{x}\right)(x+4)=80

\implies (80-x)(x+4)=80x \implies 80x-x^2+320-4x=80x \implies x^2+4x-320=0 \implies x^2+20x-16x-320 \implies x(x+20)-16(x+20)=0 \implies (x+20)(x-16)=0 \implies x+20=0 \ or\ x-16=0 \implies x=-20 \ , \ x=16 \because x\neq-20 \implies x=16 \therefore Number of books he bought = 16


Let the original duration of the tour be \ x \ days. Number of days \ \times \ expenses for each day = 360 \implies \ Expenses for each day = \displaystyle\frac{360}{x} If number of days are increased by 4, the expenses for each day is decreased by Rs.3. \implies \ Number of days = x+4 And Expenses for each day = \left(\displaystyle\frac{360}{x}-3\right) Number of days \ \times \ expenses for each day = 360 \implies (x+4)\left(\displaystyle\frac{360}{x}-3\right)=360 \implies \displaystyle\frac{(x+4)(360-3x)}{x}=360 \implies 360x+1440-12x-3x^2=360x \implies 3x^2+12x-1440=0 \implies 3x^2+72x-60x-1440=0 \implies 3x(x+24)-60(x+24)=0 \implies (x+24)(3x-60)=0 \implies x+24=0 \ (or) \ 3x-60=0 \implies \displaystyle x=-24 \ , \ x=\frac{60}{3}=20

\because \ number of days can’t be negative \ \implies x \neq -24 \therefore Original duration of the tour = 20 days .


Let the number of students be \ x \ . Number of students \ \times\ money each student got = 1200 \implies \ Money each student got = \displaystyle \frac{1200}{x} If number of students increased by ‘8’ , then the money each student got is decreased by‘5' . i.e, number of students = x+8 Money each student got = \displaystyle \frac{1200}{x}-5 Number of students \ \times \ money each student got = 1200 \implies \displaystyle (x+8) \left(\frac{1200}{x}-5\right)=1200 \implies \displaystyle\frac{(x+8)(1200-5x)}{x}=1200 \implies \displaystyle\frac{(x+8)(1200-5x)}{x}=1200 \implies 1200x-5x^2+9600-40x=1200x \implies 5x^2+40x-9600=0 \implies x^2+8x-1920=0 \implies x^2+48x-40x-1920=0 \implies x(x+48)-40(x+48)=0 \implies (x+48)(x-40)=0 \implies x+48=0 \ \ (or) \ \ x-40=0 \because \ number of students can’t be negative \ \implies x\neq -48 \implies x-40=0 \implies x=40 \therefore \ number of students = 40


Let the original price of the toy be Rs.x . Number of toys \ \times \ cost of each toy = 360 Number of toys = \displaystyle\frac{360}{x} If the cost of each toy is reduced by Rs.2 , then he can buy 2 more toys . i.e., number of toys = \displaystyle \frac{360}{x}+2 Cost of each toy = x-2 Number of toys \ \times \ cost of each toy = 360 \implies \displaystyle \left(\frac{360}{x}+2\right)(x-2)=360 \implies \displaystyle \frac{(360+2x)(x-2)}{x}=360 \implies 360x+2x^2-720-4x=360x \implies 2x^2-4x-720=0 \implies x^2-2x-360=0 \implies x^2-20x+18x-360=0 \implies x(x-20)+18(x-20)=0 \implies (x-20)(x+18)=0 \implies x-20=0 \ \ (or) \ \ x+18=0 \implies x=20,x=-18

\because \ cost of a toy can’t be negative \ \implies x \neq -18 Hence , the original price of each toy = Rs.20.


Given that selling price of the toy = Rs.24 Let the cost price of the toy be Rs.x Profit percent = x% We know that, Selling price = cost price + profit \implies \ 24 = cost price +x\% \ of \ x \implies 24=\displaystyle x+\frac{x}{100} \times x \implies \displaystyle24=\frac{100x+x^2}{100} \implies x^2+100x-2400 = 0 \implies x^2+120x-20x-2400=0 \implies x(x+120)-20(x+120)=0 \implies (x+120)(x-20)=0 \implies x+120=0 \ \ (or) \ \ x-20=0 \implies x=-120 \ , \ x=20 \because \ cost price of a toy can’t be negative \ \implies x \neq -120 Hence, cost price of the toy = Rs.20


Let the original number of persons be \ x . Number of persons \ \times \ money that each got = 9000 x\ \times money that each person got = 9000 Money that each person got = \displaystyle\frac{9000}{x} If the number of persons increased by ‘15’ then the money that person each got isdecreased by ‘30’ . i.e., number of persons = x+15 Money with each person = \displaystyle \frac{9000}{x}-30 Number of persons \times money that each person got = 9000 \implies \displaystyle (x+15) \left(\frac{9000}{x}-30\right)=9000 \implies \displaystyle\frac{(x+15)(9000-30x)}{x}=9000 \implies 9000x+135000-30x^2-450x=9000x \implies -30x^2-450x+135000=0 \implies x^2+15x-4500=0 \implies x^2+75x-60x-4500=0 \implies x(x+75)-60(x+75)=0 \implies (x+75)(x-60)=0=0 \implies x+75=0 \ , \ x-60=0 \implies x=-75 \ , \ x=60 \because \ number of people can’t be negative \ \implies x\neq -75 \therefore \ number of people =60


Let the number of students be \ x \ . Number of students \ \times \ number of apples each got = 300 \implies \ Number of apples each got = \displaystyle\frac{300}{x}\ If the number of students increased by ‘10’ then each one is getting one apple less . i.e., Number of students = x+10 Number of apples that each got = \displaystyle\frac{300}{x}-1 Number of apples \ \times \ number of students = 300 \implies \displaystyle \left(\frac{300}{x}-1\right) \ \times \ (x+10)=300 \implies \displaystyle\frac{(300-x)(x+10)}{x}=300 \implies 300x-x^2+3000-10x=300x \implies -x^2-10x+3000=0 \implies x^2+10x-3000=0 \implies x^2+60x-50x-3000=0 \implies x(x+60)-50(x+60)=0 \implies (x+60)(x-50)=0 \implies x+60=0 \ \ (or) \\ x-50=0 \implies x=-60 \ , \ x=50 \because \ number of students can’t be negative \ \implies x \neq -60 Hence , the number of students = 50


Let us assume that number of students arranged in a row of square be \ x\ . i.e., Total number of students = (x \times x) + 24=x^2 + 24 \ ---------(1) If the side of a square is increased by one more student he was short of 25 students . i.e., Side of the square = x+1 Total number of students = (x+1)^2-25---------(2) From (1) and (2) we have , x^2+24=(x+1)^2-25 \implies x^2+24=x^2+2x+1-25 \implies 24=2x-24 \implies 2x=48 \implies x=24 If \ x=24 \ then total number of students = x^2+24 =24^2+24=576+24=600 Hence , the total number of students = 600


Let the total number of students be \ x \ . Number of students \times cost of food for each = 480 \implies x \ \times cost of food for each = 480 Cost of food for each = \displaystyle\frac{480}{x} If ‘8’ members were not present then cost of food for each is increased by Rs.10. i.e., Number of students = x-8

Cost of food for each = \displaystyle\frac{480}{x}+10 Number of students \ \times\ cost of food for each = 480 \implies \displaystyle (x-8) \left(\frac{480}{x}+10\right) = 480 \implies (x-8) \displaystyle\left(\frac{480+10x}{x}\right) \implies \displaystyle\frac{(x-8)(480+10x)}{x}=480 \implies 480x+10x^2-3840-80x=480x \implies 10x^2-80x-3840=0 \implies x^2-8x-384=0 \implies x^2-24x+16x-384=0 \implies x(x-24)+16(x-24)=0 \implies (x-24)(x+16)=0 \implies x-24=0 \ \ (or) \ \ x+16=0 \implies x=24 \ , \ x=-16 \because \ number of students can’t be negative \ \implies x\neq-16 Number of students attended the picnic =Total number of students-number of studentsabsent = 24 - 8 = 16 Hence , the number of students attended the picnic = 16.


The given equation is \ x^2+5kx+16=0\ . Comparing with \ ax^2+bx+c=0\ , we have a=1 \ , \ b=5k \ , \ c=16 Discriminant \ D=b^2-4ac =(5k)^2-4(1)(16) =25k^2-64 Given that the equation has no real roots. i.e., \ \ D < 0 \implies 25k^2-64<0 \implies (5k)^2-8^2<0 \implies -8<5k<8 \implies \displaystyle -\frac{8}{5} < k < \frac{8}{5} Hence , the given equation has no real roots for \ k\ if \ \displaystyle -\frac{8}{5} < k <\frac{8}{5}


Let the numerator of the fraction be \ x\ . Then , Denominator = 2x+1 \implies Fraction = \displaystyle \frac{x}{2x+1} \implies Reciprocal of the fraction = \displaystyle \frac{2x+1}{x} Given that the sum of the fraction and its reciprocal is \ \displaystyle 2\frac{16}{21} \implies \displaystyle \frac{x}{2x+1}+\frac{2x+1}{x}=2\frac{16}{21} \implies \displaystyle \frac{x^2+(2x+1)^2}{x(2x+1)}=\frac{58}{21} \implies 21(x^2+4x^2+4x+1)=58(2x^2+x) \implies 105x^2+84x+21=116x^2+58x \implies 116x^2-105x^2+58x-84x-21=0 \implies 11x^2-26x-21=0

\implies 11x^2-33x+7x-21=0 \implies 11x(x-3)+7(x-3)=0 \implies (x-3)(11x+7)=0 \implies x-3=0 \ \ (or) \ \ 11x+7=0 \implies \displaystyle x=3\ , \ x=-\frac{7}{11} \implies x=3\ , \ x \neq \displaystyle-\frac{7}{11}\ (\because \ x \ is a natural number \\therefore x > 0 ) Hence , the required fraction = \displaystyle \frac{x}{2x+1} = \frac{3}{2(3)+1}=\frac{3}{7}


Quadratic equation roots are -2, \displaystyle \frac{1}{5} is

x^2 - (-2 + \frac{1}{5})x + (-2)\frac{1}{5} = 0

\displaystyle x^2 - (- \frac{9}{5}x ) - \frac{-2}{5} = 0

\displaystyle \frac{x^2}{1} + \frac{9x}{5} - \frac{2}{5} = 0

\displaystyle \frac{5x^2 +9x -2}{5}=0

5x^2+9x-2 = 0 \times 5=0

By comparing 5x^2 + px + q = 0

p=9 , q = -2


Let the denominator be \ x \implies numerator =x-2 \implies the original fraction = \displaystyle \frac{x-2}{x} If one is added to both numerator and denominator the new fraction obtained is \displaystyle \frac{x-2+1}{x+1} \implies \displaystyle =\frac{x-1}{x+1} Given that the sum of new and original fractions is \ \displaystyle \frac{19}{15} \implies \displaystyle \frac{x-2}{x}+\frac{x-1}{x+1}=\frac{19}{15} \implies \displaystyle \frac{(x-2)(x+1)+x(x-1)}{x(x+1)}=\frac{19}{15} \implies \displaystyle \frac{x^2-x-2+x^2-x}{x^2+x}=\frac{19}{15} \implies \displaystyle \frac{2x^2-2x-2}{x^2+x}=\frac{19}{15} \implies 15(2x^2-2x-2)=19(x^2+x) \implies 30x^2-30x-30-19x^2-19x=0 \implies 11x^2-49x-30=0 \implies 11x^2-55x+6x-30=0 \implies (x-5)(11x+6)=0 \implies x-5=0 \ \ (or) \ \ 11x+6=0 \implies x=5 \ , \ x \neq \displaystyle\frac{-6}{11} \therefore \ The original fraction is \displaystyle \frac{x-2}{x}

\displaystyle =\frac{5-2}{5}=\frac{3}{5}


Let the original price of each item be Rs.\ x\ . Price of each item \times number of items = 600 \implies number of items = \displaystyle \frac{600}{x}\ --------(1) If he keeps 10 items for himself and sold the remaining for a profit of Rs.5 per each item .

The amount received = \displaystyle \left(\frac{600}{x}-10\right)(x+5)-600

\displaystyle = \frac{(600-10x)(x+5)}{x}-600

\displaystyle = \frac{600x+3000-10x^2-50x-600x}{x}

\displaystyle = -\frac{(10x^2+50x-3000)}{x}

He can buy ‘15’ more items with the amount received in this deal.

Amount received = x \times 15

\implies \displaystyle -\frac{(10x^2+50x-3000)}{x}=15x

\implies -10x^2-50x+3000=15x^2 \implies 0=15x^2+10x^2+50x-3000 \implies 25x^2+50x-3000=0 \implies x^2+2x-120=0 \implies x^2+12x-10x-120=0 \implies (x+12)(x-10)=0 \implies x+12=0 \ , \ x-10=0 \implies x=-12 \ , \ x=10 \because \ price can’t be negative \ x \neq -12 Hence , the cost of each item Rs.10 .


Given quadratic equation is \ (p-q)x^2+5(p+q)x-2(p-q)=0. Comparing it with \ ax^2+bx+c=0\ , we have a=p-q\ ,\ b=5(p+q)\ ,\ c=-2(p-q) We know that Discriminant \ D = b^2-4ac =(5(p+q))^2-4(p-q)(-2(p-q)) =25(p+q)^2+8(p-q)^2 Given that \ p \neq q \ ,\ p\ ,\ q \ are real numbers. And we also know that the square ofany real number is posititive. \implies 25(p+q)^2+8(p-q)^2 > 0 \implies D > 0 \therefore \ The roots of given quadratic equation are real and distinct.


Given equation is \displaystyle 9\left(2x-\frac{1}{x}\right)^2-6\left(2x-\frac{1}{x}\right)+1=0

\implies \displaystyle \left[3\left(2x-\frac{1}{x}\right)\right]^2-2 \times \left[3\left(2x-\frac{1}{x}\right)\right] \times 1+1^2=0

\implies \displaystyle \left[3\left(2x-\frac{1}{x}\right)-1\right]^2=0

\implies \displaystyle 6x-\frac{3}{x}-1=0

\implies \displaystyle \frac{6x^2-x-3}{x}=0

\implies 6x^2-x-3=0 Comparing the above equation with \ ax^2+bx+c=0\ , we have a=6 \ , \ b=-1 \ , \ c=-3 We know that , D=b^2-4ac=(-1)^2-4(6)(-3)=1+72=73 By quadratic formula , we have \displaystyle x=\frac{-b\pm\sqrt{D}}{2a}=\frac{-(-1)\pm \sqrt{73}}{2(6)}=\frac{1\pm\sqrt{73}}{12}


Let x = \sqrt{2+\sqrt{2+\sqrt{2+ }}}--

Squaring on both sides x^2 = 2+ \sqrt{2+\sqrt{2+ }}--

x^2 = 2+ x

x^2 - x- 2= 0

(x - 2 ) (x + 1) = 0

x - 2=0\ \ ; x + 1 = 0

x = 2 \ \ ; x = -1

\therefore x = 2


Let us assume x = \ \sqrt{8+\sqrt{8+\sqrt{8}}}------\infty

Squaring on both sides,we get \implies x^2=8+\sqrt{8+\sqrt{8}}------\infty

\implies x^2=8+x \implies x^2-x-8=0 Comparing the above equation with \ ax^2+bx+c=0\ , we have a=1 \ , \ b=-1 \ , \ c=-8 We know that Discriminant \ D=b^2-4ac =(-1)^2-4(1)(-8)=1+32=33 > 0 By quadratic formula , we have \displaystyle x=\frac{-b\pm\sqrt{D}}{2a}=\frac{-(-1)\pm\sqrt{33}}{2(1)}

\displaystyle =\frac{1\pm\sqrt{33}}{2}


Given equation is \displaystyle x^2+\left(\frac{a}{a+b}+\frac{a+b}{a}\right)x+1=0

x^2 + \displaystyle\left(\frac{a}{a+b} + \frac{a+b}{a}\right)x + \frac{a}{a+b}\times\frac{a+b}{a}=0

A quadratic equation x^2 + (\alpha + \beta)x + \alpha\beta = 0 has roots x = -\alpha, x = -\beta

x = \displaystyle\frac{-a}{a+b}, \frac{-(a+b)}{a}


Given equation is \ \displaystyle \frac{a}{x-b}+\frac{b}{x-a}=2

\implies \displaystyle\frac{a(x-a)+b(x-b)}{(x-b)(x-a)}=2

\implies ax-a^2+bx-b^2=2(x^2-ax-bx+ab) \implies 0=2x^2-2ax-2bx+2ab-ax+a^2+b^2-bx \implies 2x^2-3ax-3bx+a^2+b^2+2ab=0 \implies 2x^2-3x(a+b)+(a+b)^2=0 \implies 2x^2-2x(a+b)-x(a+b)+(a+b)^2=0 \implies 2x(x-(a+b))-(a+b)(x-(a+b))=0 \implies 2x-(a+b))(x-(a+b))=0 \implies 2x-(a+b)=0 \ \ (or) \ x-(a+b)=0 \displaystyle x=\frac{(a+b)}{2} \ \ (or) \ \ a+b


Given that the roots of \ p(q-r)x^2+q(r-p)x+r(p-q)=0\ are equal

In ax^2+bx+c = 0, a+b+c = 0 \ roots \ \ are \ \ 1, \frac{c}{a}

Here p(q-r) + q(r-p) + r(p-q) = pq - pr + qr - pq + rp - rq = 0

so roots are 1, \displaystyle\frac{r(p-q)}{p(q-r)}

By hypothesis 1 = \displaystyle\frac{r(p-q)}{p(q-r)} ( As roots are equal according to thegiven problem)

p(q-r) = r(p-q)

pq-pr = rp -rq

pq +rq = pr + pr

pq+rq = 2pr

Divide pqr an bothsides

\displaystyle\frac{pq}{pqr} +\frac{rq}{pqr} = \frac{2pr}{pqr}

\displaystyle\frac{1}{r} + \frac{1}{p} = \frac{2}{q}


The given equation is \ \sqrt{2x+1}+\sqrt{3x+2}=\sqrt{5x+3}

Squaring on both sides of the above equation , we have

(\sqrt{2x+1})^2+(\sqrt{3x+2})^2+2\sqrt{2x+1}.\sqrt{3x+2} =(\sqrt{5x+3})^2 [ \because (a+b)^2 = a^2 + b^2 + 2ab] \implies 2x+1+3x+2+2\sqrt{(2x+1)(3x+2)}=5x+3 \implies 5x+3+2\sqrt{(2x+1)(3x+2)}=5x+3 \implies 2\sqrt{(3x+2)(2x+1)}=0 \implies (3x+2)(2x+1)=0 \implies 2x+1=0 \ \ (or) \ \ 3x+2=0 \implies \displaystyle x=-\frac{1}{2} \ , \ x=-\frac{2}{3}


Given equation is \ 3^{1+x}+3^{1-x}=10 \implies 3.3^x+\displaystyle\frac{3}{3^x}=10 Let \ 3^x=a \implies \displaystyle 3a+\frac{3}{a}=10 \implies \displaystyle \frac{3a^2+3}{a}=10 \implies 3a^2-10a+3=0 \implies 3a^2-9a-a+3=0 \implies 3a(a-3)-1(a-3)=0 \implies (3a-1)(a-3)=0

\implies 3a-1=0 \ \ (or) \ \ a-3=0 \implies \displaystyle a=\frac{1}{3} \ , \ a=3

\implies 3^x=3^{-1} \ , \ 3^x=3^1

x=-1,x=1

Hence x=-1,1.


Given that , \alpha,\beta \ are the roots of \ ax^2+bx+c=0 So, we have \ \displaystyle \alpha +\beta=\frac{-b}{a},\alpha \beta= \frac{c}{a}

Now , \ \displaystyle \frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha \beta}=\frac{(\alpha+\beta)^2-2\alpha \beta}{\alpha \beta}

\displaystyle =\frac{\left(\displaystyle \frac{-b}{a}\right)^2-2\left(\displaystyle\frac{c}{a}\right)}{\displaystyle \left(\frac{c}{a}\right)}

\displaystyle =\frac{\displaystyle \left(\frac{b^2-2ac}{a^2}\right)}{\displaystyle \left(\frac{c}{a}\right)}

\displaystyle =\frac{b^2-2ac}{ac}

Also , \ \displaystyle \frac{\alpha}{\beta}.\frac{\beta}{\alpha}=\frac{\alpha \beta}{\alpha\beta}=1

Now the required quadratic equation is given by , x^2-(sum \ of \ the \ roots )x+product \ of\ roots=0 \implies \displaystyle x^2-\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)x+\frac{\alpha}{\beta}.\frac{\beta}{\alpha}

\implies \displaystyle x^2-\left(\frac{b^2-2ac}{ac}\right)x+1=0

\implies acx^2-(b^2-2ac)x+ac=0


Given equation is \ \sqrt{5x+9}-\sqrt{7x+14}=\sqrt{2x+5} \implies \sqrt{5x+9}-\sqrt{2x+5}=\sqrt{7x+14} Squaring on both sides of the above equation , [ \because (a-b)^2 = a^2 + b^2 - 2ab] (\sqrt{5x+9})^2+(\sqrt{2x+5})^2-2\sqrt{(5x+9)(2x+5)}=(\sqrt{7x+14})^2 \implies 5x+9+2x+5-2\sqrt{(5x+9)(2x+5)}=7x+14 \implies 7x+14-2\sqrt{(5x+9)(2x+5)}=7x+14 \implies -2\sqrt{(5x+9)(2x+5)}=0 \implies (5x+9)(2x+5)=0

\implies 5x+9=0 \ \ (or) \ \ 2x+5=0 \implies \displaystyle x=\frac{-9}{5} \ \ (or) \ \ \frac{-5}{2}


x^2 + bx +c = 0 and

x^2 + cx + b = 0 have a common root then

b + c+ 1 = 0

b+ c = -1

b + c + 9 = -1 + 9 = 8


Given equation is \displaystyle \left(x+\frac{1}{x}\right)^2-\frac{3}{2}\left(x-\frac{1}{x}\right)=4

\implies \displaystyle \left(x-\frac{1}{x}\right)^2+4 (x)(\frac{1}{x})-\frac{3}{2}\left(x-\frac{1}{x}\right)=4

\implies \displaystyle 2\left(x-\frac{1}{x}\right)^2-3\left(x-\frac{1}{x}\right)+8=8

\implies \displaystyle 2\left(x-\frac{1}{x}\right)^2-3\left(x-\frac{1}{x}\right)=0

\implies \displaystyle \left(x-\frac{1}{x}\right)\left[2\left(x-\frac{1}{x}\right)-3\right]=0

\implies \displaystyle x-\frac{1}{x}=0 \hspace{1cm}or \implies \displaystyle2\left(x-\frac{1}{x}\right)-3=0 \implies x=\displaystyle\frac{1}{x} \hspace{2.5cm} \implies \displaystyle x-\frac{1}{x}=\frac{3}{2} \implies x^2=1 \hspace{2.4cm}\implies \displaystyle \frac{x^2-1}{x}=\frac{3}{2} \implies x=\pm1 \hspace{2.3cm} \implies 2x^2-2=3x \hspace{2.2cm}\implies 2x^2-3x-2=0 \hspace{2.2cm}\implies (2x+1)(x-2)=0 \hspace{2cm} \implies x=\displaystyle-\frac{1}{2}\ ,\ x=2 Hence , the value of ‘x’ in given equation is \ x=\pm1\ ,\ \displaystyle -\frac{1}{2} \ , \ 2


Given quadratic equation is \ (c^2-ab)x^2-2(a^2-bc)x+(b^2-ac)=0\ -------(1) Comparing the above equation with \ Ax^2+Bx+C=0\ , we have A=c^2-ab \ , \ B=-2(a^2-bc) \ , \ C=b^2-ac Given that equation (1) has equal roots. \implies \ Discriminant \ D=0

\implies B^2-4AC=0 \implies (-2(a^2-bc))^2-4(c^2-ab)(b^2-ac)=0 \implies \small{4(a^4+b^2c^2-2a^2bc)-4(b^2c^2-ac^3-ab^3+a^2bc)=0} \implies \small{4a^4+4b^2c^2-8a^2bc-4b^2c^2+4ac^3+4ab^3-4a^2bc=0} \implies 4a^4+4ac^3+4ab^3-12a^2bc=0 \implies 4a(a^3+b^3+c^3-3abc)=0 \implies 4a=0 \hspace{2cm}(or)\hspace{1.2cm}a^3+b^3+c^3-3abc=0 \implies a=0 \hspace{2.5cm}(or) \hspace{1.2cm}a^3+b^3+c^3=3abc Hence , \ a^3+b^3+c^3=3abc\


Given equation is \ 7^{1-x}+7^{1+x}=50 \implies \displaystyle\frac{7}{7^x}+7(7^x)=50 Let \ 7^x=a \implies \displaystyle \frac{7}{a}+7a=50 \implies 7+7a^2=50a \implies 7a^2-50a+7=0 \implies 7a^2-49a-a+7=0 \implies 7a(a-7)-1(a-7)=0 \implies (7a-1)(a-7)=0 \implies 7a-1=0 \ \ \ ,\hspace{2cm}a-7=0 \implies \displaystyle a=\frac{1}{7}\ \ \ ,\hspace{3.5cm}a=7 \implies 7^x=7^{-1}\ \ \ (or) \ \ \ 7^x=7^1 x=-1\ \ \ (or) \ \ x=1 x=\pm1


Given equation is \ \displaystyle x^{\frac{2}{3}}+x^{\frac{1}{3}}-2=0 \implies \displaystyle \left(x^{\frac{1}{3}}\right)^2+x^{\frac{1}{3}}-2=0 Let \ \displaystyle x^{\frac{1}{3}}=a \implies a^2+a-2=0 \implies (a+2)(a-1)=0 \implies a+2=0\ \ \ (or) \ \ \ a-1=0 \implies a=-2 \ \ \ (or) \ \ \ a=1 \implies \displaystyle x^{\frac{1}{3}}=-2 \ \ \ (or) \ \ \ x^{\frac{1}{3}}=1 Cubing on both sides , we have \implies \displaystyle \left(x^{\frac{1}{3}}\right)^3=(-2)^3 \hspace{2cm}\left(x^{\frac{1}{3}}\right)^3=1^3 \implies x=-8 \hspace{4cm} x=1 \implies x=1,-8


Given equation is \ (x-a)(x-b)=c^2 \implies x^2-(a+b)x+ab-c^2=0

Comparing the above equation with \ Ax^2+Bx+C=0\ , we have A=1 \ ,\ B=-(a+b)\ ,\ C=ab-c^2 Discriminant \ D=B^2-4AC =[-(a+b)]^2-4(1)(ab-c^2) =a^2+b^2+2ab-4ab+4c^2 =a^2+b^2-2ab+4c^2 =(a-b)^2+4c^2 >0 \ (\because (a-b)^2 >0 \ and \ c^2>0) \implies D > 0 Hence , the roots of \ (x-a)(x-b)=c^2\ are real and distinct .


Total money spent for wheat = Rs.120 = 12000 paise Let the cost of wheat per kg = x\ paise x \times number of kgs of wheat purchased =12000 Number of kgs of wheat purchased = \displaystyle \frac{12000}{x}\\overline{\hspace{2cm}} \ (1) If the cost is increased by 10 paise per kg , then (x+10) \times number of kgs = 12000 Number of kgs of wheat at increased price = \displaystyle \frac{12000}{x+10} \\overline{\hspace{2cm}} \ (2) According to hypothesis ,

\displaystyle \frac{12000}{x}-\frac{12000}{x+10}=5

\implies \displaystyle 12000\left(\frac{x+10-x}{x(x+10)}\right)=5

\implies \displaystyle 12000 \times \frac{10}{x^2+10x}=5 \implies 24000=x^2+10x \implies x^2+10x-24000=0 \implies x^2+160x-150x-24000=0 \implies (x+160)(x-150)=0 \implies x+160=0 \ \ \ (or) \ \ \ x-150=0 \implies x=-160 \ \ \ (or) \ \ \ x=150 As price can’t be negative , the original price of wheat is 150 paise per kg . \therefore New price of wheat = x+10 = 150+10 = 160\ paise per kg


Given that \ x^2-px+1=0\ has no real roots , then Discriminant D < 0 \implies b^2-4ac<0 \implies (-p)^2-4\times1\times1<0 \implies p^2-4<0 \implies p^2<4 \implies -2 < p < 2


Given equation is \displaystyle \frac{a}{ax-1}+\frac{b}{bx-1}=a+b

\implies \displaystyle \frac{a(bx-1)+b(ax-1)}{(ax-1)(bx-1)}=a+b

\small{ \implies abx-a+abx-b=(a+b)(abx^2-ax-bx+1)} \small{ \implies 2abx-a-b=(a+b)(abx^2-(a+b)x+1)} \implies \small{ 0=(a+b)(abx^2-(a+b)x+1)-2abx+a+b} \implies \small{ 0=x^2(a+b)ab-x((a+b)^2+2ab)+a+b+a+b} \implies \small{ 0=x^2(a+b)ab-x(a+b)^2-2abx+2(a+b)} \small{ \implies (a+b)x(abx-(a+b))-2(abx-(a+b))=0} \implies [abx-(a+b)][(a+b)x-2]=0 \implies abx-(a+b)=0\ \ \ (or) \ \ \ (a+b)x-2=0 \implies \displaystyle x=\frac{a+b}{ab} \ or \ x=\frac{2}{a+b}


If ax^2 + bx + c = 0 \ such \ that \ a+b+c = 0 then roots are 1 and \displaystyle\frac{c}{a}

In 3x^2 + 7x – 10 = 0

Sum of the coeffiecents = 3 + 7-10 = 0

\implies Roots are 1, \displaystyle\frac{-10}{3}\hspace{1cm} [ \because 1, \frac{c}{a} arethe roots]


Given equation is pm² + (\sqrt{3}-\sqrt{2})m-1 = 0

⇒ p\displaystyle\left(\frac{1}{\sqrt{3}}\right)^2 + (\sqrt{3}-\sqrt{2})\frac{1}{\sqrt{3}}-1 = 0

p\displaystyle\frac{1}{3} + \frac{\sqrt{3}}{\sqrt{3}} - \frac{\sqrt{2}}{\sqrt{3}}-1 = 0

\displaystyle\frac{p}{3}- \frac{\sqrt{2}}{\sqrt{3}} = 0

\displaystyle\frac{p}{3} = \frac{\sqrt{2}}{\sqrt{3}}

\therefore p = \sqrt{6}


In ax² + bx + c = 0

sum of the roots = \displaystyle\frac{-b}{a} Sum of roots of given equation is = \frac{-1}{1} = -1


x = 3 +2\sqrt3

x - 3 = 2\sqrt3

Squaring on both sides ,we get

(x-3)² = (2\sqrt{3})²

x^2 + 9 - 6x = 4(3)

x^2 - 6x + 9 -12 = 0

x^2 - 6x - 3 = 0 is the required equation


x² + kx + 12 = 0

Given α - β = 1

\implies (\alpha - \beta ) = \displaystyle\frac{\sqrt{ b^2-4ac}}{a} =1

\displaystyle = \frac{\sqrt{k^2-4(1)(12)}}{(1)} = 1

\sqrt {k^2-48} = 1

Squaring on both sides , we get

k² - 48 = 1

k² = 1+ 48 = 49 \therefore k = \sqrt{49}=\pm7


Given 2x² - kx + k = 0 has equal roots If ax^2 + bx + c = 0 has equal roots , then b²- 4ac = 0

\implies k² - 4(2)(k) = 0

k² = 8k

\therefore k = 8


Let Roots of the quadratic equation be α = 5 +2 {\sqrt6}\ \ \ β = 5 -2\sqrt6 Sum of the roots , α + β = 5 + 2{\sqrt6} + 5 -2 {\sqrt6} = 10

product of roots , α β = = (5 + 2{\sqrt6}) (5 - 2{\sqrt6}) = 5² - (2{\sqrt6})^2=25 - 24 = 1 The required equation is x² - sum of roots (x) + product of roots = 0 x^2-10x + 1 = 0


Given 2 is the root of x² + bx + 12 = 0

\implies 2^2 + b(2) + 12 = 0

4 + 2b + 12 = 0

2b = -16

b = -8

x² + bx + q= x² -8x + q= 0

Discriminent ∆ = 0 (Equal \ roots)

b² - 4ac = 0

\implies (-8)² - 4(1) (q) = 0

64 = 4q

q = \displaystyle\frac{64}{4} = 16

\therefore q = 16


If f(x) = x² + ax + b \ and \ g(x) = x² + bx + a

If (x-1) is the common factor

\implies f(1) = 1² + a(1) + b = 0 \ and \ g(1) = 1² + b(1) + a = 0

\implies 1 + a + b = 0 \ and \ 1 + a + b = 0


Given (x + 4) (x-4) = 9

x² - 16 = 9

x² = 25

x = \sqrt{25}= \pm5


Given 2, 3 are the roots of 3x² + 2kx + 2m = 0

put x = 2

\implies 3(2)² + 2k(2) + 2m = 0

12 + 4k + 2m = 0

4k + 2m = -12

2k + m = -6(1)

put x = 3

\implies 3(3)² + 2k(3) + 2m = 0

27 + 6k + 2m = 0

6k + 2m = -27

4k +2m=-12

on subtracting 2k = -15

k = \displaystyle\frac{-15}{2}

2\displaystyle\left(\frac{-15}{2}\right)+m = -6

m = -6 + 15 = 9

km = \displaystyle\frac{-15}{2}\times 9

=\displaystyle\frac {-135}{2}


If ax² + bx + c = 0 has roots in the ratio m:n , then (mn) b² = (m+n)² ac

Here , x² + px + 12 = 0 has roots in the ratio 1:3

\implies 1 \times 3 p^2 = (1+3)^2 (1) (12)

3 p^2 = 16(12)

p^2 = \displaystyle \frac{16(12)}{3} = 64 p = \sqrt{64} = \pm8


put x = 1

8^{1+1} + 8^{1-1} = 65

8^2 + 8^{0}= 65

64 + 1 = 65

\therefore x = 1

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