RS Aggarwal Solutions Class 10 Ex 10E

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 = \(2x^{ 2 }\)

Answers)

(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 \(\frac{ -2b }{ 3a }\)


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With the human population increasing at a tremendous rate, identify the reason due to which we need to manage our resources carefully?