# RS Aggarwal Class 10 Solutions Chapter 10 - Quadratic Equations Ex 10E ( 10.5)

## RS Aggarwal Class 10 Chapter 10 - Quadratic Equations Ex 10E ( 10.5) Solutions Free PDF

Find the discriminant of each of the following equations:

Q1) (i) $2x^{ 2 } – 7x + 6 = 0$ (ii) $3x^{ 2 } – 2x + 8 = 0$

(iii) $2x^{ 2 } – 5 \sqrt{ 2 } x + 4 = 0$ (iv) $\sqrt{ 3 } x ^{ 2 } + 2 \sqrt{ 2 } x – 2 \sqrt{ 3 } = 0$ (v) (x – 1)(2x – 1) = 0 (vi) 1 – x = $2x^{ 2 }$

(i) $2x^{ 2 } – 7x + 6 = 0$

Here,

a = 2,

b = – 7,

c = 6

Discriminant D is diven by:

D = $b ^{ 2 } – 4ac$

= $( – 7) ^{ 2 } – 4 \times 2 \times 6$

= 49 – 48

= 1

(ii) $3x ^{ 2 } – 2x + 8 = 0$

Here,

a = 3,

b = -2,

c = 8

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

= $( -2 ) ^{ 2 } – 4 \times 3 \times 8$

= 4 – 96

= – 92

(iii) $2x^{ 2 } – 5 \sqrt{ 2 } x + 4 = 0$

Here,

a = 2,

b = $-5 \sqrt{ 2 }$

c = 4

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

= $( – 5 \sqrt{ 2 } ) ^{ 2 } – 4 \times 2 \times 4$

= ( 25 * 2 ) – 32

= 50 – 32

= 18

(iv) $\sqrt{ 3 } x ^{ 2 } + 2 \sqrt{ 2 } x – 2 \sqrt{ 3 } = 0$

Here,

a = $\sqrt{ 3 }$

b = $2 \sqrt{ 2 }$

c = $- 2 \sqrt{ 3 }$

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

= $( 2 \sqrt{ 2 } ) ^{ 2 } – 4 \times \sqrt{ 3 } \times ( -2 \sqrt{ 3 })$

= ( 4 * 2 ) + ( 8 * 3 )

= 8 + 24

= 32

(v) (x – 1)(2x – 1) = 0

$\Rightarrow 2x ^{ 2 } – 3x + 1 = 0$

Comparing it with $ax ^{ 2 } + bx + c = 0$ we get

a = 2, b = -3 and c = 1

Therefore,

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

= $( – 3 ) ^{ 2 } – 4 \times 2 \times 1 = 9 – 8 = 1$

(vi) 1 – x = $2x ^{ 2 }$

$\Rightarrow 2x ^{ 2 } + x – 1 = 0$

a = 2,

b = 1,

c = -1

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

= $1 ^{ 2 } – 4 \times 2 ( -1 )$

= 1 + 8

= 9

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

Q2) $x ^{ 2 } – 4x – 1$

Ans. 2) Given :

$x ^{ 2 } – 4x – 1$

Comparing it with $ax ^{ 2 } + bx + c = 0$ we get:

a = 1, b = -4 and c = -1

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

$( -4 ) ^{ 2 } – 4 \times 1 \times ( -1 )$

= 16 + 4

= 20

= 20 > 0

Hence, the roots of the equation are real.

Roots $\alpha$ and $\beta$ are given by :

$\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

= $\frac{ -( -4 ) + \sqrt{ 20 } }{ 2 \times 1 }$

= $\frac{ 4 + 2 \sqrt{ 5 } }{ 2 }$

= $\frac{ 2 ( 2 + \sqrt{ 5 } ) }{ 2 } = ( 2 + \sqrt{ 5 })$

$\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

= $\frac{ – ( -4 ) – \sqrt{ 20 } }{ 2 \times 1 }$

= $\frac{ 4 – 2 \sqrt{ 5 } }{ 2 }$

= $\frac{ 2 ( 2 – \sqrt{ 5 } ) }{ 2 } = ( 2 – \sqrt{ 5 })$

Thus, the roots of the equation are $( 2 + \sqrt{ 5 })$ and $( 2 + \sqrt{ 5 })$

Q3) $x ^{ 2 } – 6x + 4 = 0$

Ans. 3) Given :

$x ^{ 2 } – 6x + 4 = 0$

On Comparing it with $ax ^{ 2 } + bx + c = 0$ we get:

a = 1, b = -6 and c =4

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

= $( – 6) ^{ 2 } – 4 \times 1 \times 4$

= 36 – 16

= 20 > 0

Hence, the roots of the equation are real.

Roots $\alpha$ and $\beta$ are given by :

$\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

= $\frac{ – ( -6 ) – \sqrt{ 20 } }{ 2 \times 1 }$

= $\frac{ 6 + 2 \sqrt{ 5 } }{ 2 }$

= $\frac{ 2 ( 3 + \sqrt{ 5 } ) }{ 2 } = ( 3 + \sqrt{ 5 })$

= $\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

= $\frac{ – ( -6 ) – \sqrt{ 20 } }{ 2 \times 1 }$

= $\frac{ 6 – 2 \sqrt{ 5 } }{ 2 }$

= $\frac{ 2 ( 2 – \sqrt{ 5 } ) }{ 2 } = ( 3 – \sqrt{ 5 })$

Thus, the roots of the equation are $( 3 + 2 \sqrt{ 5 })$ and $( 3 – 2 \sqrt{ 5 })$

Q4) $2x ^{ 2 } + x – 4 = 0$

Ans. 4) The given equation is $2x ^{ 2 } + x – 4 = 0$

On Comparing it with $ax ^{ 2 } + bx + c = 0$ we get:

a =2, b = 1 and c = -4

Therefore,

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

= $( 1 )^{ 2 } – 4 \times 2 \times ( -4 )$

= 1 + 32

= 33 > 0

So, the given equation has real roots.

Now, $\sqrt{ D }$ = $\sqrt{ 33 }$

$\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

= $\frac{ -1 + \sqrt{ 33 } }{ 2 \times 2} = \frac{ -1 + \sqrt{ 33 } }{ 4 }$

$\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

= $\frac{ -1 – \sqrt{ 33 } }{ 2 \times 2} = \frac{ -1 – \sqrt{ 33 } }{ 4 }$

Hence, $\frac{ -1 + \sqrt{ 33 } }{ 4 }$ and $\frac{ -1 – \sqrt{ 33 } }{ 4 }$ are the roots of the given equation.

Q5) $25x ^{ 2 } + 30x + 7 = 0$

Ans. 5) Given:

$25x ^{ 2 } + 30x + 7 = 0$

On Comparing it with $ax ^{ 2 } + bx + c = 0$ we get:

a = 25, b = 30 and c = 7

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

= $30^{ 2 } – 4 \times 25 \times 7$

= 900 – 700

= 200

= 200 > 0

Hence, the roots of the equation are real.

Roots $\alpha$ and $\beta$ are given by :

$\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

= $\frac{ -30 + \sqrt{ 200 } }{ 2 \times 25} = \frac{ -30 + 10 \sqrt{ 2 } }{ 50 }$

= $\frac{ 10 ( -3 + \sqrt{ 2 } ) }{ 50 } = \frac{ ( -3 + \sqrt{ 2 } ) }{ 5 }$

$\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

= $\frac{ -30 – \sqrt{ 200 } }{ 2 \times 25} = \frac{ -30 – 10 \sqrt{ 2 } }{ 50 }$

= $\frac{ 10 ( -3 – \sqrt{ 2 } ) }{ 50 } = \frac{ ( -3 – \sqrt{ 2 } ) }{ 5 }$

Thus, the roots of the equation are $\frac{ ( -3 + \sqrt{ 2 } ) }{ 5 }$ and $\frac{ ( -3 – \sqrt{ 2 } ) }{ 5 }$ .

Q6) $16x^{ 2 } = 24x + 1$

Ans. 6) Given:

$16x^{ 2 } = 24x + 1$

$\Rightarrow 16x^{ 2 } – 24x – 1 = 0$

On Comparing it with $ax ^{ 2 } + bx + c = 0$ we get:

a = 16, b = -24 and c = -1

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

= $( -24 )^{ 2 } – 4 \times 16 \times ( -1 )$

= 576 + ( 64 )

= 640 > 0

Hence, the roots of the equation are real,

Roots $\alpha$ and $\beta$ are given by :

$\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

= $\frac{ -( -24 + \sqrt{ 640 } }{ 2 \times 16} = \frac{ 24 + 8 \sqrt{ 10 } }{ 32 }$

= $\frac{ 8 ( 3 + \sqrt{ 10 } ) }{ 50 } = \frac{ ( 3 + \sqrt{ 10 } ) }{ 4 }$

$\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

= $\frac{ – ( -24 ) – \sqrt{ 640} }{ 2 \times 16} = \frac{ 24 – 8 \sqrt{ 10 } }{ 32 }$

= $\frac{ 8 ( 3 – \sqrt{ 10 } ) }{ 32 } = \frac{ ( 3 – \sqrt{ 10 } ) }{ 4 }$

Thus, the roots of the equation are $\frac{ ( 3 + \sqrt{ 10 } ) }{ 4 }$ and $\frac{ ( 3 – \sqrt{ 10 } ) }{ 4 }$ .

Q7) $15x^{ 2 } – 28 = x$

Ans. 7) Given:

$15x^{ 2 } – 28 = x$

$\Rightarrow 15x^{ 2 } – x – 28 = 0$

On Comparing it with $ax ^{ 2 } + bx + c = 0$ we get:

a = 15, b = -1 and c = -28

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

= $( -1 )^{ 2 } – 4 \times 15 \times ( -28 )$

= 1 – (- 1680)

= 1 + 1680

= 1681

= 1681 > 0

Hence, the roots of the equation are real,

Roots $\alpha$ and $\beta$ are given by :

$\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

$\frac{ -( -1) + \sqrt{ 1681 } }{ 2 \times 15} = \frac{ 1 + 41 }{ 30 }$

= $\frac{ 42 }{ 30 } = \frac{ 7 }{ 5 }$

$\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

= $\frac{ – ( -1 ) – \sqrt{ 1681 }}{ 2 \times 15 } = \frac{ 1 – 41 }{ 30 }$

= $\frac{ -40 }{ 30 } = \frac{ -4 }{ 3 }$

Thus, the roots of the equation are $\frac{ 7 }{ 5 }$ and $\frac{ -4 }{ 3 }$ .

Q8) $2x ^{ 2 } – 2 \sqrt{ 2 } x + 1 = 0$

Ans. 8) Given:

$2x ^{ 2 } – 2 \sqrt{ 2 } x + 1 = 0$

On Comparing it with $ax ^{ 2 } + bx + c = 0$ we get:

a = 2, b = $- 2 \sqrt{ 2 }$ and c = 1

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

= $(- 2 \sqrt{ 2 } ) ^{ 2 } – 4 \times 2 \times 1$

= 8 – 8 = 0

So, the given equation has real roots.

Now, $\sqrt{ D } = 0$

$\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

= $\frac{ -(-2\sqrt{ 2 } ) + \sqrt{ 0 } }{ 2 \times 2} = \frac{ 2\sqrt{ 2 }}{ 4 } =\frac{ \sqrt{ 2 } }{ 2 }$

$\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

= $\frac{ – ( – 2\sqrt{ 2 } ) – \sqrt{ 0 } }{ 2 \times 2} = \frac{ 2\sqrt{ 2 }}{ 4 } = \frac{ \sqrt{ 2 } }{ 2 }$

Hence, $\frac{ \sqrt{ 2 } }{ 2 }$ is the repeated root of the equation.

Q9) $\sqrt{ 2 } x^{ 2 } + 7x + 5 \sqrt{ 2 } = 0$

Ans. 9) The given equation is $\sqrt{ 2 } x^{ 2 } + 7x + 5 \sqrt{ 2 } = 0$

On Comparing it with $ax ^{ 2 } + bx + c = 0$ we get:

a = $\sqrt{ 2 }$ , b = 7 and c = $5 \sqrt{ 2 }$

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

= $( 7 ) ^{ 2 } – 4 \times \sqrt{ 2 } \times 5 \sqrt{ 2 }$

= 49 – 40 = 9 > 0

So, the given equation has real roots.

Now, $\sqrt{ D } = \sqrt{ 9 } = 3$

Therefore, $\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

$\frac{ -7 + 3 }{ 2 \times \sqrt{ 2 } } = \frac{ -4 }{ 2 \sqrt{ 2 } } = -\sqrt{ 2 }$

$\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

= $\frac{ -7 – 3 }{ 2 \times \sqrt{ 2 } } = \frac{ -10 }{ 2 \sqrt{ 2 } } = – \frac{ 5 \sqrt{ 2 } }{ 2 }$

Hence, $- \sqrt{ 2 }$ and $- \frac{ 5 \sqrt{ 2 } }{ 2 }$ are the roots of the given equation.

Q10) $\sqrt{ 3 } x^{ 2 } + 10x + 8 \sqrt{ 3 } = 0$

Ans. 10) Given:

$\sqrt{ 2 } x^{ 2 } + 7x + 5 \sqrt{ 2 } = 0$

On Comparing it with $ax ^{ 2 } + bx + c = 0$ we get:

a = $\sqrt{ 3 }$ , b = 10 and c = $-8 \sqrt{ 3 }$

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

= $( 10 )^{ 2 } – 4 \times \sqrt{ 3 } \times ( -8 \sqrt{ 3 })$

= 100 + 96

= 196 > 0

Hence, the roots of the equation are real.

Roots $\alpha$ and $\beta$ are given by :

$\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

= $\frac{ -10 + \sqrt{ 196 } }{ 2 \sqrt{ 3 }} = \frac{ -10 +14 }{ 2 \sqrt{ 3 }}$

= $\frac{ 4 }{ 2 \sqrt{ 3 }} = \frac{ 2 }{ \sqrt{ 3 } }$

= $\frac{ 2 }{ \sqrt{ 3 }} \times \frac{ \sqrt{ 3 }}{ \sqrt{ 3 }} = \frac{ 2 \sqrt{ 3 }}{ 3 }$

$\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

= $\frac{ -10 – \sqrt{ 196 } }{ 2 \sqrt{ 3 }} = \frac{ -10 – 14 }{ 2 \sqrt{ 3 }}$

= $\frac{ -24 }{ 2 \sqrt{ 3 }} = \frac{ -12 }{ \sqrt{ 3 } }$

= $\frac{ -12 }{ \sqrt{ 3 }} \times \frac{ \sqrt{ 3 }}{ \sqrt{ 3 }} = \frac{ -12 \sqrt{ 3 }}{ 3 }$

= $-4 \sqrt{ 3 }$

Thus, the roots of the equation are $\frac{ 2 \sqrt{ 3 }}{ 3 }$ and $-4 \sqrt{ 3 }$

Q11) $\sqrt{ 3 } x^{ 2 } – 2 \sqrt{ 2 }x – 2 \sqrt{ 3 } = 0$

Ans. 11) The given equation is $\sqrt{ 3 } x^{ 2 } – 2 \sqrt{ 2 }x – 2 \sqrt{ 3 } = 0$

On Comparing it with $ax ^{ 2 } + bx + c = 0$ we get:

a = $\sqrt{ 3 }$ , b = $- 2 \sqrt{ 2 }$ and c = $- 2 \sqrt{ 3 }$

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

$( -2 \sqrt{ 2 } )^{ 2 } – 4 \times \sqrt{ 3 } \times ( -2 \sqrt{ 3 })$

= 8 + 24 = 32 > 0

So, the given equation has real roots.

Now, $\sqrt{ D } = \sqrt{ 32 } = 4 \sqrt{ 2 }$

$\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

= $\frac{ -(-2 \sqrt{ 2 } ) + 4 \sqrt{ 2 } }{2 \times \sqrt{ 3 }}$

= $\frac{ 6 \sqrt{ 2 } }{ 2 \sqrt{ 3 } } = \sqrt{ 6 }$

$\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

= $\frac{ -(-2 \sqrt{ 2 } ) – 4 \sqrt{ 2 } }{2 \times \sqrt{ 3 }}$

= $\frac{ -2 \sqrt{ 2 } }{ 2 \sqrt{ 3 } } = -\frac{ \sqrt{ 6 } }{ 3 }$

Hence, $\sqrt{ 6 }$ and $\frac{ – \sqrt{ 6 } }{ 3 }$ are the roots of the given equation.

Q12) $2x ^{ 2 } + 6 \sqrt{ 3 } x – 60 = 0$

Ans. 12) The given equation is $2x ^{ 2 } + 6 \sqrt{ 3 } x – 60 = 0$

On Comparing it with $ax ^{ 2 } + bx + c = 0$ we get:

a = 2, b = $6 \sqrt{ 3 }$ and c = -60

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

= $(6 \sqrt{ 3 })^{ 2 } – 4 \times 2 \times ( -60 )$

= 108 + 480

= 588 > 0

So, the given equations has real roots.

Now, $\sqrt{ D } = \sqrt{ 588 } = 14 \sqrt{ 3 }$

$\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

= $\frac{ -6 \sqrt{ 3 } + 14 \sqrt{ 3 }}{ 2 \times 2} = \frac{ 8 \sqrt{ 3 }}{ 4 } = 2 \sqrt{ 3 }$

$\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

= $\frac{ -6 \sqrt{ 3 } – 14 \sqrt{ 3 }}{ 2 \times 2} = \frac{ -20 \sqrt{ 3 }}{ 4 } = -5 \sqrt{ 3 }$

Hence, $2 \sqrt{ 3 }$ and $-5 \sqrt{ 3 }$ are the roots of the given equation.

Q13) $4 \sqrt{ 3 } x^{ 2 } + 5x – 2\sqrt{ 3 } = 0$

Ans. 13) The given equation is $4 \sqrt{ 3 } x^{ 2 } + 5x – 2\sqrt{ 3 } = 0$

On Comparing it with $ax ^{ 2 } + bx + c = 0$ we get:

a = $4 \sqrt{ 3 }$ , b = 5 and c = $-2 \sqrt{ 3 }$

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

$5^{ 2 } – 4 \times 4 \sqrt{ 3 } \times ( -2 \sqrt{ 3 })$

25 + 96 = 121 > 0

So, the given equation has real roots.

Now, $\sqrt{ D } = \sqrt{ 121 }$ =11

Therefore, $\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

$\frac{ -5 + 11 }{ 2 \times 4 \sqrt{ 3 } } = \frac{ 6 }{ 8 \sqrt{ 3 } } = \frac{ \sqrt{ 3 } }{ 4 }$

$\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

$\frac{ -5 – 11 }{ 2 \times 4 \sqrt{ 3 } } = \frac{ -16 }{ 8 \sqrt{ 3 } } = \frac{ 2 \sqrt{ 3 } }{ 3 }$

Hence, $\frac{ \sqrt{ 3 } }{ 4 }$ and $\frac{ 2 \sqrt{ 3 } }{ 3 }$ are the roots of the given equation.

Q14) $3x ^{ 2 } – 2 \sqrt{ 6 } x + 2 = 0$

Ans. 14) The given equation is $3x ^{ 2 } – 2 \sqrt{ 6 } x + 2 = 0$

On Comparing it with $ax ^{ 2 } + bx + c = 0$ we get:

a = 3, b = $- 2 \sqrt{ 6 }$ and c = 2

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

$( – 2 \sqrt{ 6 } )^{ 2 } – 4 \times 3 \times 2$

= 24 – 24

= 0

So, the given equation has real roots.

Now, $\sqrt{ D } = 0$

Therefore, $\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

$\frac{ -( -2 \sqrt{ 6} ) + 0 }{ 2\times 3} = \frac{ 2 \sqrt{ 6 } }{ 6 } = \frac{ \sqrt{ 6 } }{ 3 }$

$\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

$\frac{ -( -2 \sqrt{ 6} ) – 0 }{ 2\times 3} = \frac{ 2 \sqrt{ 6 } }{ 6 } = \frac{ \sqrt{ 6 } }{ 3 }$

Hence, $\frac{ \sqrt{ 6 } }{ 3 }$ is the repeated root of the given equation.

Q15) $2 \sqrt{ 3 } x^{ 2 } – 5x + \sqrt{ 3 } = 0$

Ans. 15) The given equation is $2 \sqrt{ 3 } x^{ 2 } – 5x + \sqrt{ 3 } = 0$

On Comparing it with $ax ^{ 2 } + bx + c = 0$ we get:

a = $2 \sqrt{ 3 }$ , b = -5 and c = $\sqrt{ 3 }$

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

Therefore, $( -5 )^{ 2 } – 4 \times 2\sqrt{ 3 } \times \sqrt{ 3 }$

25 – 24 = 1 > 0

So, the given equation has real roots.

Now, $\sqrt{ D } = \sqrt{ 1 } = 1$

Therefore, $\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

$\frac{ -( -5 ) + 1 }{ 2 \times 2 \sqrt{ 3 }} = \frac{ 6 }{ 4 \sqrt{ 3 } } = \frac{ \sqrt{ 3 } }{ 2 }$

$\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

$\frac{ -( -5 ) – 1 }{ 2 \times 2 \sqrt{ 3 }} = \frac{ 4 }{ 4 \sqrt{ 3 } } = \frac{ \sqrt{ 3 } }{ 3 }$

Hence, $\frac{ \sqrt{ 3 } }{ 2 }$ and $\frac{ \sqrt{ 3 } }{ 3 }$ are the roots of the given equation.

Q16) $x^{ 2 } + x + 2 = 0$

Ans. 16) The given equation is $x^{ 2 } + x + 2 = 0$

On Comparing it with $ax ^{ 2 } + bx + c = 0$ we get:

a = 1, b = 1 and c = 2

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

= $1^{ 2 } – 4 \times 1 \times 2$

= 1 – 8 = – 7 < 0

Hence, the given equation has no real roots (or real roots does not exist).

Q17) $2x^{ 2 } + ax – a^{ 2 } = 0$

Ans. 17) The given equation is $2x^{ 2 } + ax – a^{ 2 } = 0$

On Comparing it with $Ax ^{ 2 } + Bx + C = 0$ we get:

A = 2, B = a and C = $- a^{ 2 }$

Discriminant D is given by:

D = $B ^{ 2 } – 4AC$

$a^{ 2 } – 4 \times 2 \times -a^{ 2 } = a^{ 2 } + 8a^{ 2 } = 9a^{ 2 } \geq 0$

So, the given equation has real roots.

Now, $\sqrt{ D } = \sqrt{ 9 a^{ 2 } } = 3a$

Therefore, $\alpha$ = $\frac{ -B + \sqrt{ D } } { 2A }$

$\frac{ -a + 3a }{ 2 \times 2} = \frac{ 2a }{ 4 } = \frac{ a }{ 2 }$

$\beta$ = $\frac{ -B – \sqrt{ D } } { 2A }$

$\frac{ -a – 3a }{ 2 \times 2} = \frac{ -4a }{ 4 } = -a$

Hence, $\frac{ a }{ 2 }$ and -a are the roots of the given equation.

Q18) $x^{ 2 } – ( \sqrt{ 3 } +1 )x + \sqrt{ 3 } = 0$

Ans. 18) The given equation is $x^{ 2 } – ( \sqrt{ 3 } +1 )x + \sqrt{ 3 } = 0$

On Comparing it with $ax ^{ 2 } + bx + c = 0$ we get:

a = 1, b = – ( $\sqrt{ 3 } +1$ ) and c = $\sqrt{ 3 }$

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

$\sqrt{ D } = \left [-( \sqrt{ 3 } – 1 ) \right ]^{ 2 } – 4 \times 1 \times \sqrt{ 3 }$

= 3 + 1 + $2 \sqrt{ 3 } – 4 \sqrt{ 3 }$ = 3 – $2 \sqrt{ 3 }$ + 1

= $( \sqrt{ 3 } – 1 )^{ 2 }$ > 0

So, the given equation has real roots.

Now, $\sqrt{ D } = \sqrt{( \sqrt{ 3 } – 1 )^{ 2 }} = \sqrt{ 3 } – 1$

Therefore, $\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

= $\frac{ -\left [ -(\sqrt{ 3 } + 1 ) \right ] + ( \sqrt{ 3 } -1 ) }{2 \times 1}$

= $\frac{ \sqrt{ 3 } + 1 + \sqrt{ 3 } – 1}{ 2 } = \frac{ 2 \sqrt{ 3 } }{ 2 } = \sqrt{ 3 }$

$\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

= $\frac{ -\left [ -(\sqrt{ 3 } + 1 ) \right ] – ( \sqrt{ 3 } -1 ) }{2 \times 1}$

= $\frac{ \sqrt{ 3 } + 1 – \sqrt{ 3 } + 1 }{ 2 } = \frac{ 2 }{ 2 } = 1$

Hence, $\sqrt{ 3 }$ and 1 are the roots of the given equation.

Q19) $2x^{ 2 } + 5 \sqrt{ 3 } x + 6 = 0$

Ans. 19) The given equation is $2x^{ 2 } + 5 \sqrt{ 3 } x + 6 = 0$

On Comparing it with $ax ^{ 2 } + bx + c = 0$ we get:

a = 2, b = $5 \sqrt{ 3 }$ and c = 6

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

= $\left (5 \sqrt{ 3 } \right ) ^{ 2 } – 4 \times 2 \times 6$

75 – 48 = 27 > 0

So, the given equation has real roots.

Now, $\sqrt{ D } = \sqrt{ 27 } = 3 \sqrt{ 3 }$

Therefore, $\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

= $\frac{ -5 \sqrt{ 3 } + 3 \sqrt{ 3 } }{ 2\times 2} = \frac{ -2 \sqrt{ 3 } }{ 4 } =- \frac{ \sqrt{ 3 } }{ 2}$

$\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

= $\frac{ -5 \sqrt{ 3 } – 3 \sqrt{ 3 } }{ 2 \times 2} = \frac{ -8 \sqrt{ 3 } }{ 4 } = -2 \sqrt{ 3 }$

Hence, $- \frac{ \sqrt{ 3 } }{ 2}$ and $-2 \sqrt{ 3 }$ are the roots of the given equation.

Q20) $3x^{ 2 } – 2x + 2 = 0$

Ans. 20) The given equation is$3x^{ 2 } – 2x + 2 = 0$

On Comparing it with $ax ^{ 2 } + bx + c = 0$ we get:

a = 3, b = -2 and c = 2

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

= $(-2)^{ 2 } – 4 \times 3 \times 2$

= 4 – 24 = -20 < 0

Hence, the given equation has no real roots (or real roots does not exist).

Q21) x + $\frac{ 1 }{ x } = 3 , x \neq 0$

Ans. 21) The given equation is x + $\frac{ 1 }{ x } = 3 , x \neq 0$

$\frac{ x ^{ 2 } + 1 }{ x } = 3$

$x^{ 2 } + 1 = 3x$

$x^{ 2 } – 3x + 1 = 0$

This equation is of the form $ax ^{ 2 } + bx + c = 0$ , where a = 1, b = -3 and c = 1

Discriminant D is given by:

D = $b ^{ 2 } – 4ac$

= $( -3 )^{ 2 } – 4 \times 1 \times 1$

= 9 – 4 = 5 > 0

So, the given equation has real roots.

Now, $\sqrt{ 5 } = \sqrt{ 5 }$

Therefore, $\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

= $\frac{ -( -3 ) + \sqrt{ 5 } }{ 2 \times 1} = \frac{ 3 + \sqrt{ 5 } }{ 2 }$

$\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

= $\frac{ -( -3 ) – \sqrt{ 5 } }{ 2 \times 1} = \frac{ 3 – \sqrt{ 5 } }{ 2 }$

Hence, $\frac{ 3 + \sqrt{ 5 } }{ 2 }$ and $\frac{ 3 – \sqrt{ 5 } }{ 2 }$ are the roots of the given equation.

Q22) $\frac{ 1 }{ x } – \frac{ 1 }{ x – 2 } = 3$

Ans. 22) The given equation is $\frac{ 1 }{ x } – \frac{ 1 }{ x – 2 } = 3, \; x\neq 0 , 2$

$\frac{ x – 2 – x}{ x(x – 2)} = 3$

$\frac{ -2 }{ ( x^{ 2 } – 2x )} = 3$

$-2 = 3x^{ 2 } – 6x$

$3x^{ 2 } – 6x + 2 = 0$

This equation is of the form $ax ^{ 2 } + bx + c = 0$ , where a = 3, b = -6 and c = 2

Discriminant D is given by:

D = $b^{ 2 } – 4ac$

= $( -6 )^{ 2 } – 4 \times 3 \times 2$

= 36 – 24 = 12 > 0

So, the given equation has real roots.

Now, $\sqrt{ D } = \sqrt{ 12 } = 2 \sqrt{ 3 }$

Therefore, $\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

= $\frac{ -( -6 ) + 2 \sqrt{ 3 } }{ 2 \times 3} = \frac{ 6 + 2 \sqrt{ 3 } }{ 6 } = \frac{ 3 +\sqrt{ 3 }}{ 3 }$

$\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

= $\frac{ -( -6 ) – 2 \sqrt{ 3 } }{ 2 \times 3} = \frac{ 6 – 2 \sqrt{ 3 } }{ 6 } = \frac{ 3 – \sqrt{ 3 }}{ 3 }$

Hence, $\frac{ 3 + \sqrt{ 3 } }{ 3 }$ and $\frac{ 3 – \sqrt{ 3 } }{ 3 }$ are the roots of the given equation.

Q23) $x – \frac{ 1 }{ x } = 3, \; x \neq 0$

Ans. 23) The given equation is

$x – \frac{ 1 }{ x } = 3, \; x \neq 0$

$\frac{ x^{ 2 } – 1 }{ x } = 3$

$x^{ 2 } – 1 = 3x$

$x^{ 2 } – 3x – 1 = 0$

This equation is of the form $ax ^{ 2 } + bx + c = 0$

a = 1, b = -3 and c = -1.

Discriminant D is given by:

D = $b^{ 2 } – 4ac$

= $( -3 )^{ 2 } – 4 \times 1 \times ( -1 )$

= 9 + 4 = 13 > 0

So, the given equation has real roots.

Now, $\sqrt{ D } = \sqrt{ 13 }$

Therefore, $\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

$\frac{ -( -3 ) + \sqrt{ 13 } }{ 2 \times 1} = \frac{ 3 + \sqrt{ 13 } }{ 2 }$

$\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

$\frac{ -( -3 ) – \sqrt{ 13 } }{ 2 \times 1} = \frac{ 3 – \sqrt{ 13 } }{ 2 }$

Hence, $\frac{ 3 + \sqrt{ 13 } }{ 2 }$ and $\frac{ 3 – \sqrt{ 13 } }{ 2 }$ are the roots of the given equation.

Q24) $\frac{ m }{ n }x^{ 2 } + \frac{ n }{ m } = 1 -2x$

Ans. 24) The given equation is

$\frac{ m }{ n }x^{ 2 } + \frac{ n }{ m } = 1 -2x$

$\frac{ m^{ 2 } x^{ 2 } + n^{ 2 } }{ mn } = 1 – 2x$

$m^{ 2 } x^{ 2 } + n^{ 2 }$ = mn – 2mnx

$m^{ 2 } x^{ 2 } + 2mnx + n^{ 2 } – mn = 0$

This equation is of the form $ax ^{ 2 } + bx + c = 0$ , where a = $m^{ 2 }$ , b = 2mn and c = $n^{ 2 }$ – mn

Discriminant D is given by:

D = $b^{ 2 } – 4ac$

= $( 2mn )^{ 2 } – 4 \times m^{ 2 } \times ( n^{ 2 } – mn ) = 4m^{ 2 } n^{ 2 } – 4m^{ 2 }n^{ 2 } + 4m^{ 3 }n = 4m^{ 3 }n > 0$

So, the given equation has real roots

Now, $\sqrt{ D } = \sqrt{ 4m^{ 3 } n } = 2m \sqrt{ mn }$

Therefore, $\alpha$ = $\frac{ -b + \sqrt{ D } } { 2a }$

$\frac{ -2mn + 2m \sqrt{ mn } }{ 2 \times m^{ 2 } } = \frac{ 2m (-n + \sqrt{ mn } ) }{ 2m^{ 2 }} = \frac{ -n + \sqrt{ mn } }{ m }$

$\beta$ = $\frac{ -b – \sqrt{ D } } { 2a }$

$\frac{ -2mn – 2m \sqrt{ mn } }{ 2 \times m^{ 2 }} = \frac{ -2m (-n + \sqrt{ mn } ) }{ 2m^{ 2 }} = \frac{ -n – \sqrt{ mn } }{ m }$

Hence, $\frac{ -n + \sqrt{ mn } }{ m }$ and $\frac{ -n – \sqrt{ mn } }{ m }$ are the roots of the given equation.

Q25) $36x^{ 2 } -12ax + ( a^{ 2 } – b^{ 2 } ) = 0$

Ans. 25) The given equation is $36x^{ 2 } -12ax + ( a^{ 2 } – b^{ 2 } ) = 0$

On Comparing it with $Ax ^{ 2 } + Bx + C = 0$ we get:

A = 36, B = -12a and C = $a^{ 2 } – b^{ 2 }$

Discriminant D is given by:

D = $B^{ 2 } – 4AC$

$( -12a )^{ 2 } – 4 \times 36 \times ( a^{ 2 } – b^{ 2 } ) = 144a^{ 2 } – 144a^{ 2 } + 144b^{ 2 } = 144b^{ 2 } > 0$

So, the given equation has real roots.

Now, $144b^{ 2 } = 12b$

Therefore, $\alpha$ = $\frac{ -B + \sqrt{ D } } { 2A }$

$\frac{ -( -12a ) + 12b }{ 2 \times 36} = \frac{ 12 (a + b) }{ 72 } = \frac{ a + b }{ 6 }$

$\beta$ = $\frac{ -B – \sqrt{ D } } { 2A }$

$\frac{ -( -12a ) – 12b }{ 2 \times 36} = \frac{ 12 (a – b) }{ 72 } = \frac{ a – b }{ 6 }$

Hence, $\frac{ a + b }{ 6 }$ and $\frac{ a – b }{ 6 }$ are the roots of the given equation.

Q26) $x^{ 2 } – 2ax + ( a^{ 2 } – b^{ 2 } ) = 0$

Ans. 26) Given:

$x^{ 2 } – 2ax + ( a^{ 2 } – b^{ 2 } ) = 0$

On Comparing it with $Ax ^{ 2 } + Bx + C = 0$ we get:

A = 1, B = -2a and C = $( a^{ 2 } – b^{ 2 } )$

Discriminant D is given by:

D = $B^{ 2 } – 4AC$

= $( -2a )^{ 2 } – 4 \times 1 \times ( a^{ 2 } – b^{ 2 } )$

= $4a^{ 2 } – 4a^{ 2 } + 4b^{ 2 }$

= $4b^{ 2 } > 0$

Hence, the roots of the equation are real.

Roots $\alpha$ and $\beta$ are given by :

Therefore, $\alpha$ = $\frac{ -B + \sqrt{ D } } { 2A }$

= $\frac{ -(-2a ) + \sqrt{ 4b^{ 2 } } }{ 2\times 1}$

= $\frac{ 2a + 2b }{ 2 } = \frac{ 2( a +b ) }{ 2 } = ( a + b )$

$\beta$ = $\frac{ -B – \sqrt{ D } } { 2A }$

= $\frac{ -(-2a ) – \sqrt{ 4b^{ 2 } } }{ 2\times 1}$

= $\frac{ 2a – 2b }{ 2 } = \frac{ 2( a – b ) }{ 2 } = ( a – b )$

Hence, the roots of the equation are ( a + b) and ( a – b ).

Q27) $x^{ 2 } – 2ax – ( 4b^{ 2} – a^{ 2 } ) = 0$

Ans. 27) The given equation is $x^{ 2 } – 2ax – ( 4b^{ 2} – a^{ 2 } ) = 0$

On Comparing it with $Ax ^{ 2 } + Bx + C = 0$ we get:

A = 1, B = -2a and C = $- ( 4b^{ 2} – a^{ 2 } )$

Discriminant D is given by:

D = $B^{ 2 } – 4AC$

= $( -2a )^{ 2 } – 4 \times 1 \times \left [ -( 4b^{ 2 } – a^{ 2 }) \right ]$

= $4a^{ 2 } + 16b^{ 2 } – 4a^{ 2 } = 16b^{ 2 } > 0$

So, the given equation has real roots.

Now, $\sqrt{ D } = \sqrt{ 16b^{ 2 } } = 4b$

Therefore, $\alpha$ = $\frac{ -B + \sqrt{ D } } { 2A }$

= $\frac{ -( -2a ) + 4b }{ 2 \times 1} = \frac{ 2( a + 2b )}{ 2 } = a + 2b$

$\beta$ = $\frac{ -B – \sqrt{ D } } { 2A }$

= $\frac{ -( -2a ) – 4b }{ 2 \times 1} = \frac{ 2( a – 2b ) }{ 2 } = a – 2b$

Hence, a + 2b and a – 2b are the roots of the given equation.

Q28) $x^{ 2 } + 6x – ( a^{ 2 } + 3a – 8 ) = 0$

Ans. 28) The given equation is $x^{ 2 } + 6x – ( a^{ 2 } + 3a – 8 ) = 0$

On Comparing it with $Ax ^{ 2 } + Bx + C = 0$ we get:

A = 1 , B = 6 and C = $-( a^{ 2 } + 3a – 8 )$

Discriminant D is given by:

D = $B^{ 2 } – 4AC$

= $6^{ 2 } – 4 \times 1 \times \left [ -(a^{ 2 } + 2a – 8 ) \right ]$

= 36 + $4a^{ 2 }$ + 8a – 32 = $4a^{ 2 }$ + 8a + 4

= $4( a^{ 2 } + 2a + 1 ) = 4( a + 1 )^{ 2 } > 0$

So, the given equation has real roots.

Now, $\sqrt{ D } = \sqrt{ 4(a + 1)^{ 2 } }$ = 2 (a + 1)

Therefore, $\alpha$ = $\frac{ -B + \sqrt{ D } } { 2A }$

= $\frac{ -6 + 2(a + 1) }{ 2 \times 1} = \frac{ 2a – 4 }{ 2 } = a – 2$

$\beta$ = $\frac{ -B – \sqrt{ D } } { 2A }$

= $\frac{ -6 – 2(a + 1) }{ 2 \times 1} = \frac{ -2a – 8 }{ 2 } = -a – 2 = -( a + 4)$

Hence, (a – 2) and -(a + 4) are the roots of the given equation.

Q29) $x^{ 2 } + 5x – ( a^{ 2 } + a – 6 ) = 0$

Ans. 29) The given equation is $x^{ 2 } + 5x – ( a^{ 2 } + a – 6 ) = 0$

On Comparing it with $Ax ^{ 2 } + Bx + C = 0$ we get:

A = 1, B = 5 and C = $-( a^{ 2 } + a – 6 )$

Discriminant D is given by:

D = $B^{ 2 } – 4AC$

= $5^{ 2 } – 4 \times 1 \times \left [ -( a^{ 2 } + a – 6 ) \right ]$

= 25 + $4a^{ 2 }$ + 4a – 24 = $4a^{ 2 }$ + 4a + 1

= $( 2a + 1 )^{ 2 } > 0$

So, the given equation has real roots.

Now, $\sqrt{ D } = \sqrt{ ( 2a + 1 )^{ 2 } }$ = 2a + 1

Therefore, $\alpha$ = $\frac{ -B + \sqrt{ D } } { 2A }$

= $\frac{ -5 + 2a + 1 }{ 2 \times 1 } = \frac{ 2a – 4 }{ 2 } = a – 2$

$\beta$ = $\frac{ -B – \sqrt{ D } } { 2A }$

= $\frac{ -5 – ( 2a + 1 ) }{ 2 \times 1 } = \frac{ -2a – 6 }{ 2 } = -a – 3 = -(a + 3)$

Hence, (a – 2) and -(a + 3) are the roots of the given equation.

Q30) $x^{ 2 } – 4ax – b^{ 2 } + 4a^{ 2 } = 0$

Ans. 30) The given equation is $x^{ 2 } – 4ax – b^{ 2 } + 4a^{ 2 } = 0$

On Comparing it with $Ax ^{ 2 } + Bx + C = 0$ we get:

A = 1, B = -4a and C = $-b^{ 2 } + 4a^{ 2 }$

Discriminant D is given by:

D = $B^{ 2 } – 4AC$

= $( -4a )^{ 2 } – 4 \times 1 \times ( -b^{ 2 } + 4a^{ 2 } )$

= $16a^{ 2 } + 4b^{ 2 } – 16a^{ 2 } = 4b^{ 2 } > 0$

So, the given equation has real roots.

Now, $\sqrt{ D } = \sqrt{ 4b^{ 2 } } = 2b$

Therefore, $\alpha$ = $\frac{ -B + \sqrt{ D } } { 2A }$

= $\frac{ -(-4a) + 2b }{ 2 \times 1} = \frac{ 4a + 2b }{ 2 } = 2a + b$

$\beta$ = $\frac{ -B – \sqrt{ D } } { 2A }$

= $\frac{ -(-4a) – 2b }{ 2 \times 1} = \frac{ 4a – 2b }{ 2 } = 2a – b$

Hence, (2a + b) and (2a – b) are the roots of the given equation.

Q31) $4x^{ 2 } – 4a^{ 2 } x + ( a^{ 4 } – b^{ 4 } ) = 0$

Ans. 31) The given equation is $4x^{ 2 } – 4a^{ 2 } x + ( a^{ 4 } – b^{ 4 } ) = 0$

On Comparing it with $Ax ^{ 2 } + Bx + C = 0$ we get:

A = 4, B = $- 4a^{ 2 }$ and C = $a^{ 4 } – b^{ 4 }$

Discriminant D is given by:

D = $B^{ 2 } – 4AC$

= $(-4a^{ 2 } )^{ 2 } – 4 \times 4 \times (a^{ 4 } – b^{ 4 } )$

= $16a^{ 4 } – 16a^{ 4 } + 16b^{ 4 } = 16b^{ 4 } > 0$

So, the given equation has real roots.

Therefore, $\alpha$ = $\frac{ -B + \sqrt{ D } } { 2A }$

= $\frac{ -(-4a ) + 4b^{ 2 } }{ 2 \times 4} = \frac{ 4( a^{ 2 } + b^{ 2 } ) }{ 8 } = \frac{ a^{ 2 } + b ^{ 2 } }{ 2 }$

$\beta$ = $\frac{ -B – \sqrt{ D } } { 2A }$

= $\frac{ -(-4a ) – 4b^{ 2 } }{ 2 \times 4} = \frac{ 4( a^{ 2 } – b^{ 2 } ) }{ 8 } = \frac{ a^{ 2 } – b ^{ 2 } }{ 2 }$

Hence, $\frac{ a^{ 2 } + b ^{ 2 } }{ 2 }$ and $\frac{ a^{ 2 } – b ^{ 2 } }{ 2 }$ are the roots of the given equation.

Q32) $4x^{ 2 } + 4bx – ( a^{ 2 } – b^{ 2 } ) = 0$

Ans. 32) The given equation is $4x^{ 2 } + 4bx – ( a^{ 2 } – b^{ 2 } ) = 0$

On Comparing it with $Ax ^{ 2 } + Bx + C = 0$ we get:

A = 4, B = 4b and C = $-( a^{ 2 } – b^{ 2 } )$

Discriminant D is given by:

D = $B^{ 2 } – 4AC$

= $\left (4b \right )^{ 2 } – 4 \times 4 \times \left [ -( a^{ 2 } – b^{ 2 } ) \right ]$

= $16b^{ 2 } + 16a ^{ 2 } – 16b^{ 2 } = 16a^{ 2 } > 0$

So, the given equation has real roots.

Now, $\sqrt{ D } = \sqrt{ 16a^{ 2 } } = 4a$

Therefore, $\alpha$ = $\frac{ -B + \sqrt{ D } } { 2A }$

= $\frac{ -4b + 4a }{ 2 \times 4 } = \frac{ 4(a – b) }{ 8 } = \frac{ a – b }{ 2 }$

$\beta$ = $\frac{ -B – \sqrt{ D } } { 2A }$

= $\frac{ -4b – 4a }{ 2 \times 4 } = \frac{ 4(a + b) }{ 8 } = \frac{ a + b }{ 2 }$

Hence, $\frac{ a – b }{ 2 }$ and $-\frac{ a + b }{ 2 }$ are the roots of the given equation.

Q33) $x^{ 2 } – (2b – 1)x + (b^{ 2 } – b – 20) = 0$

Ans. 33) The given equation is $x^{ 2 } – (2b – 1)x + (b^{ 2 } – b – 20) = 0$

On Comparing it with $Ax ^{ 2 } + Bx + C = 0$ we get:

A = 1, B = – ( 2b – 1 ) and C = $b^{ 2 } – b – 20$

Discriminant D is given by:

D = $B^{ 2 } – 4AC$

= $\left [ -( 2b – 1 ) \right ]^{ 2 } – 4 \times 1 \times ( b^{ 2 } – b – 20 )$

= $4b^{ 2 } – 4b + 1 – 4b^{ 2 } + 4b + 80 = 81 > 0$

So, the given equation has real roots.

Now, $\sqrt{ D } = \sqrt{ 81 } = 9$

Therefore, $\alpha$ = $\frac{ -B + \sqrt{ D } } { 2A }$

= $\frac{ -\left [ -(2b – 1) \right ] + 9 }{ 2\times 1 } = \frac{ 2b + 8 }{ 2 }$ = b + 4

$\beta$ = $\frac{ -B – \sqrt{ D } } { 2A }$

= $\frac{ -\left [ -(2b – 1) \right ] – 9 }{ 2\times 1 } = \frac{ 2b – 10 }{ 2 }$ = b – 5

Hence, (b + 4) and (b – 5) are the roots of the given equation.

Q34) $3a^{ 2 }x^{ 2 } + 8abx + 4b^{ 2 } = 0$

Ans. 34) Given:

$3a^{ 2 }x^{ 2 } + 8abx + 4b^{ 2 } = 0$

On Comparing it with $Ax ^{ 2 } + Bx + C = 0$ we get:

A = $3a^{ 2 }$ , B = 8ab and C = $4b^{ 2 }$

Discriminant D is given by:

D = $B^{ 2 } – 4AC$

= $( 8ab )^{ 2 } – 4 \times 3a^{ 2 } \times 4b^{ 2 }$

= $16a^{ 2 }b^{ 2 } > 0$

Hence, the roots of the equation are real.

Roots $\alpha$ and $\beta$ are given by :

Therefore, $\alpha$ = $\frac{ -B + \sqrt{ D } } { 2A }$

= $\frac{ -8ab + \sqrt{ 16a^{ 2 } b^{ 2 }}}{ 2 \times 3a^{ 2 } } = \frac{ -8ab + 4ab }{ 6a^{ 2 } }$

= $\frac{ -4ab }{ 6a^{ 2 } } = \frac{ -2b }{ 3a }$

$\beta$ = $\frac{ -B – \sqrt{ D } } { 2A }$

= $\frac{ -8ab – \sqrt{ 16a^{ 2 } b^{ 2 } } }{ 2 \times 3a^{ 2 } } = \frac{ -8ab – 4ab }{ 6a^{ 2 } }$

= $\frac{ -12ab }{ 6a^{ 2 } } = \frac{ -2b }{ a }$

Thus, the roots of the equation are $\frac{ -2b }{ 3a }$ and $\frac{ -2b }{ a }$

Q35) $a^{ 2 }b^{ 2 }x^{ 2 } – ( 4b^{ 4 } – 3a^{ 4 } )x – 12a^{ 2 }b^{ 2 } = 0$

Ans. 35) The given equation is $a^{ 2 }b^{ 2 }x^{ 2 } – ( 4b^{ 4 } – 3a^{ 4 } )x – 12a^{ 2 }b^{ 2 } = 0$

On Comparing it with $Ax ^{ 2 } + Bx + C = 0$ we get:

A = $a^{ 2 }b^{ 2 }$ , B = $-( 4b^{ 4 } – 3a^{ 4 } )$ and C = $-12a^{ 2 }b^{ 2 }$

Discriminant D is given by:

D = $B^{ 2 } – 4AC$

= $\left [ -( 4b^{ 4 } – 3a^{ 4 } ) \right ]^{ 2 } – 4 \times a^{ 2 }b^{ 2 } \times (-12a^{ 2 }b^{ 2 })$

= $16b^{ 8 } + 24a^{ 4 }b^{ 4 } + 9a^{ 8 } = ( 4b^{ 4 } + 3a^{ 4 } )^{ 2 } > 0$

So, the given equation has real roots.

Now, $\sqrt{ D } = \sqrt{ ( 4b^{ 4 } + 3a^{ 4 } )^{ 2 } } = 4b^{ 4 } + 3a^{ 4 }$

Therefore, $\alpha$ = $\frac{ -B + \sqrt{ D } } { 2A }$

= $\frac{ -\left [ -( 4b^{ 4 } – 3a^{ 4 } ) \right ] +( 4b^{ 4 } + 3a^{ 4 } ) }{ 2 \times a^{ 2 }b^{ 2 } }$

= $\frac{ 8b^{ 4 } }{ 2a^{ 2 } b^{ 2 }} = \frac{ 4b^{ 2 } }{ a^{ 2 } }$

$\beta$ = $\frac{ -B – \sqrt{ D } } { 2A }$

= $\frac{ -\left [ -( 4b^{ 4 } – 3a^{ 4 } ) \right ] – ( 4b^{ 4 } + 3a^{ 4 } ) }{ 2 \times a^{ 2 }b^{ 2 } }$

= $\frac{ – 6a^{ 4 } }{ 2a^{ 2 } b^{ 2 }} = – \frac{ 3a^{ 2 } }{ b^{ 2 } }$

Hence, $\frac{ 4b^{ 2 } }{ a^{ 2 } }$ and $- \frac{ 3a^{ 2 } }{ b^{ 2 } }$ are the roots of the given equation.

Q36) $12abx^{ 2 } – ( 9a^{ 2 } – 8b^{ 2 } )x – 6ab = 0$

Ans. 36) Given:

$12abx^{ 2 } – ( 9a^{ 2 } – 8b^{ 2 } )x – 6ab = 0$

On Comparing it with $Ax ^{ 2 } + Bx + C = 0$ we get:

A = 12ab , B = $- 9a^{ 2 } – 8b^{ 2 }$ and C = -6ab

Discriminant D is given by:

D = $B^{ 2 } – 4AC$

= $\left [ -( 9a^{ 2 } – 8b^{ 2 } ) \right ]^{ 2 } – 4 \times 12ab \times ( -6ab )$

= $81a^{ 4 } – 144a^{ 2 }b^{ 2 } + 64b^{ 2 } + 288a^{ 2 }b^{ 2 }$

= $81a^{ 4 } + 144a^{ 2 }b^{ 2 } + 64b^{ 2 }$

= $( 9a ^{ 2 } + 8b^{ 2 } )^{ 2 } > 0$

Hence, the roots of the equation are equal.

Roots $\alpha$ and $\beta$ are given by :

Therefore, $\alpha$ = $\frac{ -B + \sqrt{ D } } { 2A }$

= $\frac{ -\left [ -( 9a^{ 2 } – 8b^{ 2 } ) \right ] +\sqrt{ ( 9a^{ 2 } +8b^{ 2 }) ^{ 2 }}}{ 2 \times 12ab }$

= $\frac{ 9a^{ 2 } – 8b^{ 2 } + 9a^{ 2 } + 8b^{ 2} }{ 24ab } = \frac{ 18a^{ 2 } }{ 24ab } = \frac{ 3a }{ 4b }$

$\beta$ = $\frac{ -B – \sqrt{ D } } { 2A }$

= $\frac{ -\left [ -( 9a^{ 2 } – 8b^{ 2 } ) \right ] – \sqrt{ ( 9a^{ 2 } +8b^{ 2 }) ^{ 2 } } }{ 2 \times 12ab }$

= $\frac{ 9a^{ 2 } – 8b^{ 2 } – 9a^{ 2 } – 8b^{ 2} }{ 24ab } = \frac{ -16b^{ 2 } }{ 24ab } = \frac{ -2b }{ 3a }$

Thus, the roots of the equation are $\frac{ 3a }{ 4b }$ and $\frac{ -2b }{ 3a }$

#### Practise This Question

Which of the following is the exponential notation of 81?