QBManager

Complex Numbers and Quadratic Equation

118468 If $α, β$ are the roots of the quadratic equation $x^{2} + a x + b = 0, (b \neq 0)$ , then the quadratic equation whose roots are $α - \frac{1}{β}, β - \frac{1}{α}$ , is

1

a x^{2} + a (b - 1) x + (a - 1)^{2} = 0

2

b x^{2} + a (b - 1) x + (b - 1)^{2} = 0

3

x^{2} + a x + b v = 0

4

a b x^{2} + b x + a = 0

Explanation:

B Given,
$x^{2} + a x + b = 0 b \neq 0$
$α + β = - a$
$α \cdot β = b$
Now,
$(α - \frac{1}{β}) + (β - \frac{1}{α}) = (α + β) - (\frac{α + β}{α \cdot β})$
$= (- a) - (\frac{- a}{b})$
$= - a + \frac{a}{b}$
$= - \frac{a}{b} (1 - b)$
And,
$(α - \frac{1}{β}) (β - \frac{1}{α}) = α β - 1 - 1 + \frac{1}{α β}$
$= b + \frac{1}{b} - 2$
$= \frac{1}{b} (b^{2} - 2 b + 1) = \frac{1}{b} (b - 1)^{2}$
Required of equation whose roots are
$(α - \frac{1}{β}) and (β - \frac{1}{α}) is,$
$x^{2} - [(α - \frac{1}{β}) + (β - \frac{1}{α})] x + [(α - \frac{1}{β}) \cdot (β - \frac{1}{α})] = 0$
$x^{2} - \frac{a}{b} (1 - b) x + \frac{1}{b} (b - 1)^{2} = 0$
${bx}^{2} + a (b - 1) x + (b - 1)^{2} = 0$

WB JEE-2013

Complex Numbers and Quadratic Equation

118469 The quadratic equation whose sum of the roots is 11 and sum of squares of the roots is 61 is

1

x^{2} + 11 x - 30 = 0

2

x^{2} + 11 x + 30 = 0

3

x^{2} - 11 x - 30 = 0

4

x^{2} - 11 x + 30 = 0

Explanation:

D Sum of the roots $(α + β) = 11$
Square of the $(α^{2} + β^{2}) = 61$
We know that,
$(α + β)^{2} = (α^{2} + β^{2}) + 2 α β$
$(11)^{2} = 61 + 2 α β$
$α β = \frac{121 - 61}{2}$
$α β = 30$
The quadratic equation,
$x^{2} - (α + β) x + (α β) = 0$
$x^{2} - 11 x + 30 = 0$

AP EAMCET-08.07.2022

Complex Numbers and Quadratic Equation

118470 Let $p, q \in R$ . If $2 - \sqrt{3}$ is a root of the quadratic equation, $x^{2} + p + q = 0$ , then

1

q^{2} - 4 p - 16 = 0

2

p^{2} - 4 q - 12 = 0

3

p^{2} - 4 p + 12 = 0

4

q^{2} + 4 p + 14 = 0

Explanation:

B Given,
Let $p, q \in R$ .
If we take root as any real number $β$ , then quadratic equation will be
$x^{2} - (β + 2 - \sqrt{3}) x + β (2 - \sqrt{3}) = 0$
Now, we can have none or any of the options can be correct depending upon ' $β$ '
Instead of $p, q \in R$ it should be $p, q \in R$ it should be $p$ , $q \in Q$ then other root will be $2 + \sqrt{3}$
$\Rightarrow p = - (q + \sqrt{3} - 2 - \sqrt{3}) = - 4$
And, $q = (2 + \sqrt{3}) (2 - \sqrt{3}) = 1$
$\Rightarrow p^{2} - 4 q - 12 = (- 4)^{2} - 4 - 12 = 0$ Hence the correct (b)

JEE Main 09.04.2019

Complex Numbers and Quadratic Equation

118471 The quardratic equation whose roots are $\frac{1}{3 + \sqrt{2}}$ and $\frac{1}{3 - \sqrt{2}}$ , will be

1

7 x^{2} - 6 x + 1 = 0

2

6 x^{2} - 7 x + 1 = 0

3

x^{2} - 6 x + 7 = 0

4

x^{2} - 7 x + 6 = 0

Explanation:

A Given,
The quadratic equation whose roots are
$\frac{1}{3 + \sqrt{2}} and \frac{1}{3 - \sqrt{2}}$
So, the required quadratic equation is $x^{2}$ (sum of the roots) $x +$ product of the roots $= 0$
$x^{2} - (\frac{1}{3 + \sqrt{2}} + \frac{1}{3 - \sqrt{2}}) x + (\frac{1}{3 + \sqrt{2}} \times \frac{1}{3 - \sqrt{2}}) = 0$
$x^{2} - (\frac{3 - \sqrt{2} + 3 + \sqrt{2}}{(3 + \sqrt{2}) (3 - \sqrt{2})}) x + \frac{1}{3^{2} - (\sqrt{2})^{2}} = 0$
$x^{2} - \frac{6}{9 - 2} x + \frac{1}{9 - 2} = 0$
$x^{2} - \frac{6}{7} x + \frac{1}{7} = 0$
$7 x^{2} - 6 x + 1 = 0$

Manipal UGET-2020

Complex Numbers and Quadratic Equation

118472 If $α, β$ are the roots of $a x^{2} + b x + c = 0$ , then the quadratic equation whose roots are $\sqrt{5} α, \sqrt{5} β$ is

1

a x^{2} + \sqrt{5} b x + 5 c = 0

2

a x^{2} + \sqrt{5} bx + \sqrt{5} c = 0

3

a x^{2} + 5 b x + \sqrt{5} c = 0

4

a x^{2} + 5 bx + 5 c = 0

Explanation:

A Given,
If $α, β$ are the roots of
$a x^{2} + b x + c = 0$
$α + β = - \frac{b}{a}$
$α \cdot β = \frac{c}{a}$
If the roots of quadratic equation is $\sqrt{5} α$ and $\sqrt{5} β$ then the equation.
$x^{2} - (Sum of roots) x + (product of roots) = 0$
$\Rightarrow α = \frac{- b}{a} β = \frac{c}{a}$
$\Rightarrow x^{2} - (\sqrt{5} α + \sqrt{5} β) x + (\sqrt{5} α \times \sqrt{5} β) = 0$
$\Rightarrow x^{2} - \sqrt{5} (α + β) x + 5 α β = 0$
$\Rightarrow x^{2} - \sqrt{5} (\frac{- b}{a}) x + \frac{5 c}{a} = 0$
$\Rightarrow {ax}^{2} + \sqrt{5} b x + 5 c = 0$

AP EAMCET-07.07.2022

Complex Numbers and Quadratic Equation

118468 If $α, β$ are the roots of the quadratic equation $x^{2} + a x + b = 0, (b \neq 0)$ , then the quadratic equation whose roots are $α - \frac{1}{β}, β - \frac{1}{α}$ , is

1

a x^{2} + a (b - 1) x + (a - 1)^{2} = 0

2

b x^{2} + a (b - 1) x + (b - 1)^{2} = 0

3

x^{2} + a x + b v = 0

4

a b x^{2} + b x + a = 0

Explanation:

B Given,
$x^{2} + a x + b = 0 b \neq 0$
$α + β = - a$
$α \cdot β = b$
Now,
$(α - \frac{1}{β}) + (β - \frac{1}{α}) = (α + β) - (\frac{α + β}{α \cdot β})$
$= (- a) - (\frac{- a}{b})$
$= - a + \frac{a}{b}$
$= - \frac{a}{b} (1 - b)$
And,
$(α - \frac{1}{β}) (β - \frac{1}{α}) = α β - 1 - 1 + \frac{1}{α β}$
$= b + \frac{1}{b} - 2$
$= \frac{1}{b} (b^{2} - 2 b + 1) = \frac{1}{b} (b - 1)^{2}$
Required of equation whose roots are
$(α - \frac{1}{β}) and (β - \frac{1}{α}) is,$
$x^{2} - [(α - \frac{1}{β}) + (β - \frac{1}{α})] x + [(α - \frac{1}{β}) \cdot (β - \frac{1}{α})] = 0$
$x^{2} - \frac{a}{b} (1 - b) x + \frac{1}{b} (b - 1)^{2} = 0$
${bx}^{2} + a (b - 1) x + (b - 1)^{2} = 0$

WB JEE-2013

Complex Numbers and Quadratic Equation

118469 The quadratic equation whose sum of the roots is 11 and sum of squares of the roots is 61 is

1

x^{2} + 11 x - 30 = 0

2

x^{2} + 11 x + 30 = 0

3

x^{2} - 11 x - 30 = 0

4

x^{2} - 11 x + 30 = 0

Explanation:

D Sum of the roots $(α + β) = 11$
Square of the $(α^{2} + β^{2}) = 61$
We know that,
$(α + β)^{2} = (α^{2} + β^{2}) + 2 α β$
$(11)^{2} = 61 + 2 α β$
$α β = \frac{121 - 61}{2}$
$α β = 30$
The quadratic equation,
$x^{2} - (α + β) x + (α β) = 0$
$x^{2} - 11 x + 30 = 0$

AP EAMCET-08.07.2022

Complex Numbers and Quadratic Equation

118470 Let $p, q \in R$ . If $2 - \sqrt{3}$ is a root of the quadratic equation, $x^{2} + p + q = 0$ , then

1

q^{2} - 4 p - 16 = 0

2

p^{2} - 4 q - 12 = 0

3

p^{2} - 4 p + 12 = 0

4

q^{2} + 4 p + 14 = 0

Explanation:

B Given,
Let $p, q \in R$ .
If we take root as any real number $β$ , then quadratic equation will be
$x^{2} - (β + 2 - \sqrt{3}) x + β (2 - \sqrt{3}) = 0$
Now, we can have none or any of the options can be correct depending upon ' $β$ '
Instead of $p, q \in R$ it should be $p, q \in R$ it should be $p$ , $q \in Q$ then other root will be $2 + \sqrt{3}$
$\Rightarrow p = - (q + \sqrt{3} - 2 - \sqrt{3}) = - 4$
And, $q = (2 + \sqrt{3}) (2 - \sqrt{3}) = 1$
$\Rightarrow p^{2} - 4 q - 12 = (- 4)^{2} - 4 - 12 = 0$ Hence the correct (b)

JEE Main 09.04.2019

Complex Numbers and Quadratic Equation

118471 The quardratic equation whose roots are $\frac{1}{3 + \sqrt{2}}$ and $\frac{1}{3 - \sqrt{2}}$ , will be

1

7 x^{2} - 6 x + 1 = 0

2

6 x^{2} - 7 x + 1 = 0

3

x^{2} - 6 x + 7 = 0

4

x^{2} - 7 x + 6 = 0

Explanation:

A Given,
The quadratic equation whose roots are
$\frac{1}{3 + \sqrt{2}} and \frac{1}{3 - \sqrt{2}}$
So, the required quadratic equation is $x^{2}$ (sum of the roots) $x +$ product of the roots $= 0$
$x^{2} - (\frac{1}{3 + \sqrt{2}} + \frac{1}{3 - \sqrt{2}}) x + (\frac{1}{3 + \sqrt{2}} \times \frac{1}{3 - \sqrt{2}}) = 0$
$x^{2} - (\frac{3 - \sqrt{2} + 3 + \sqrt{2}}{(3 + \sqrt{2}) (3 - \sqrt{2})}) x + \frac{1}{3^{2} - (\sqrt{2})^{2}} = 0$
$x^{2} - \frac{6}{9 - 2} x + \frac{1}{9 - 2} = 0$
$x^{2} - \frac{6}{7} x + \frac{1}{7} = 0$
$7 x^{2} - 6 x + 1 = 0$

Manipal UGET-2020

Complex Numbers and Quadratic Equation

118472 If $α, β$ are the roots of $a x^{2} + b x + c = 0$ , then the quadratic equation whose roots are $\sqrt{5} α, \sqrt{5} β$ is

1

a x^{2} + \sqrt{5} b x + 5 c = 0

2

a x^{2} + \sqrt{5} bx + \sqrt{5} c = 0

3

a x^{2} + 5 b x + \sqrt{5} c = 0

4

a x^{2} + 5 bx + 5 c = 0

Explanation:

A Given,
If $α, β$ are the roots of
$a x^{2} + b x + c = 0$
$α + β = - \frac{b}{a}$
$α \cdot β = \frac{c}{a}$
If the roots of quadratic equation is $\sqrt{5} α$ and $\sqrt{5} β$ then the equation.
$x^{2} - (Sum of roots) x + (product of roots) = 0$
$\Rightarrow α = \frac{- b}{a} β = \frac{c}{a}$
$\Rightarrow x^{2} - (\sqrt{5} α + \sqrt{5} β) x + (\sqrt{5} α \times \sqrt{5} β) = 0$
$\Rightarrow x^{2} - \sqrt{5} (α + β) x + 5 α β = 0$
$\Rightarrow x^{2} - \sqrt{5} (\frac{- b}{a}) x + \frac{5 c}{a} = 0$
$\Rightarrow {ax}^{2} + \sqrt{5} b x + 5 c = 0$

AP EAMCET-07.07.2022

Complex Numbers and Quadratic Equation

118468 If $α, β$ are the roots of the quadratic equation $x^{2} + a x + b = 0, (b \neq 0)$ , then the quadratic equation whose roots are $α - \frac{1}{β}, β - \frac{1}{α}$ , is

1

a x^{2} + a (b - 1) x + (a - 1)^{2} = 0

2

b x^{2} + a (b - 1) x + (b - 1)^{2} = 0

3

x^{2} + a x + b v = 0

4

a b x^{2} + b x + a = 0

Explanation:

B Given,
$x^{2} + a x + b = 0 b \neq 0$
$α + β = - a$
$α \cdot β = b$
Now,
$(α - \frac{1}{β}) + (β - \frac{1}{α}) = (α + β) - (\frac{α + β}{α \cdot β})$
$= (- a) - (\frac{- a}{b})$
$= - a + \frac{a}{b}$
$= - \frac{a}{b} (1 - b)$
And,
$(α - \frac{1}{β}) (β - \frac{1}{α}) = α β - 1 - 1 + \frac{1}{α β}$
$= b + \frac{1}{b} - 2$
$= \frac{1}{b} (b^{2} - 2 b + 1) = \frac{1}{b} (b - 1)^{2}$
Required of equation whose roots are
$(α - \frac{1}{β}) and (β - \frac{1}{α}) is,$
$x^{2} - [(α - \frac{1}{β}) + (β - \frac{1}{α})] x + [(α - \frac{1}{β}) \cdot (β - \frac{1}{α})] = 0$
$x^{2} - \frac{a}{b} (1 - b) x + \frac{1}{b} (b - 1)^{2} = 0$
${bx}^{2} + a (b - 1) x + (b - 1)^{2} = 0$

WB JEE-2013

Complex Numbers and Quadratic Equation

118469 The quadratic equation whose sum of the roots is 11 and sum of squares of the roots is 61 is

1

x^{2} + 11 x - 30 = 0

2

x^{2} + 11 x + 30 = 0

3

x^{2} - 11 x - 30 = 0

4

x^{2} - 11 x + 30 = 0

Explanation:

D Sum of the roots $(α + β) = 11$
Square of the $(α^{2} + β^{2}) = 61$
We know that,
$(α + β)^{2} = (α^{2} + β^{2}) + 2 α β$
$(11)^{2} = 61 + 2 α β$
$α β = \frac{121 - 61}{2}$
$α β = 30$
The quadratic equation,
$x^{2} - (α + β) x + (α β) = 0$
$x^{2} - 11 x + 30 = 0$

AP EAMCET-08.07.2022

Complex Numbers and Quadratic Equation

118470 Let $p, q \in R$ . If $2 - \sqrt{3}$ is a root of the quadratic equation, $x^{2} + p + q = 0$ , then

1

q^{2} - 4 p - 16 = 0

2

p^{2} - 4 q - 12 = 0

3

p^{2} - 4 p + 12 = 0

4

q^{2} + 4 p + 14 = 0

Explanation:

B Given,
Let $p, q \in R$ .
If we take root as any real number $β$ , then quadratic equation will be
$x^{2} - (β + 2 - \sqrt{3}) x + β (2 - \sqrt{3}) = 0$
Now, we can have none or any of the options can be correct depending upon ' $β$ '
Instead of $p, q \in R$ it should be $p, q \in R$ it should be $p$ , $q \in Q$ then other root will be $2 + \sqrt{3}$
$\Rightarrow p = - (q + \sqrt{3} - 2 - \sqrt{3}) = - 4$
And, $q = (2 + \sqrt{3}) (2 - \sqrt{3}) = 1$
$\Rightarrow p^{2} - 4 q - 12 = (- 4)^{2} - 4 - 12 = 0$ Hence the correct (b)

JEE Main 09.04.2019

Complex Numbers and Quadratic Equation

118471 The quardratic equation whose roots are $\frac{1}{3 + \sqrt{2}}$ and $\frac{1}{3 - \sqrt{2}}$ , will be

1

7 x^{2} - 6 x + 1 = 0

2

6 x^{2} - 7 x + 1 = 0

3

x^{2} - 6 x + 7 = 0

4

x^{2} - 7 x + 6 = 0

Explanation:

A Given,
The quadratic equation whose roots are
$\frac{1}{3 + \sqrt{2}} and \frac{1}{3 - \sqrt{2}}$
So, the required quadratic equation is $x^{2}$ (sum of the roots) $x +$ product of the roots $= 0$
$x^{2} - (\frac{1}{3 + \sqrt{2}} + \frac{1}{3 - \sqrt{2}}) x + (\frac{1}{3 + \sqrt{2}} \times \frac{1}{3 - \sqrt{2}}) = 0$
$x^{2} - (\frac{3 - \sqrt{2} + 3 + \sqrt{2}}{(3 + \sqrt{2}) (3 - \sqrt{2})}) x + \frac{1}{3^{2} - (\sqrt{2})^{2}} = 0$
$x^{2} - \frac{6}{9 - 2} x + \frac{1}{9 - 2} = 0$
$x^{2} - \frac{6}{7} x + \frac{1}{7} = 0$
$7 x^{2} - 6 x + 1 = 0$

Manipal UGET-2020

Complex Numbers and Quadratic Equation

118472 If $α, β$ are the roots of $a x^{2} + b x + c = 0$ , then the quadratic equation whose roots are $\sqrt{5} α, \sqrt{5} β$ is

1

a x^{2} + \sqrt{5} b x + 5 c = 0

2

a x^{2} + \sqrt{5} bx + \sqrt{5} c = 0

3

a x^{2} + 5 b x + \sqrt{5} c = 0

4

a x^{2} + 5 bx + 5 c = 0

Explanation:

A Given,
If $α, β$ are the roots of
$a x^{2} + b x + c = 0$
$α + β = - \frac{b}{a}$
$α \cdot β = \frac{c}{a}$
If the roots of quadratic equation is $\sqrt{5} α$ and $\sqrt{5} β$ then the equation.
$x^{2} - (Sum of roots) x + (product of roots) = 0$
$\Rightarrow α = \frac{- b}{a} β = \frac{c}{a}$
$\Rightarrow x^{2} - (\sqrt{5} α + \sqrt{5} β) x + (\sqrt{5} α \times \sqrt{5} β) = 0$
$\Rightarrow x^{2} - \sqrt{5} (α + β) x + 5 α β = 0$
$\Rightarrow x^{2} - \sqrt{5} (\frac{- b}{a}) x + \frac{5 c}{a} = 0$
$\Rightarrow {ax}^{2} + \sqrt{5} b x + 5 c = 0$

AP EAMCET-07.07.2022

Complex Numbers and Quadratic Equation

118468 If $α, β$ are the roots of the quadratic equation $x^{2} + a x + b = 0, (b \neq 0)$ , then the quadratic equation whose roots are $α - \frac{1}{β}, β - \frac{1}{α}$ , is

1

a x^{2} + a (b - 1) x + (a - 1)^{2} = 0

2

b x^{2} + a (b - 1) x + (b - 1)^{2} = 0

3

x^{2} + a x + b v = 0

4

a b x^{2} + b x + a = 0

Explanation:

B Given,
$x^{2} + a x + b = 0 b \neq 0$
$α + β = - a$
$α \cdot β = b$
Now,
$(α - \frac{1}{β}) + (β - \frac{1}{α}) = (α + β) - (\frac{α + β}{α \cdot β})$
$= (- a) - (\frac{- a}{b})$
$= - a + \frac{a}{b}$
$= - \frac{a}{b} (1 - b)$
And,
$(α - \frac{1}{β}) (β - \frac{1}{α}) = α β - 1 - 1 + \frac{1}{α β}$
$= b + \frac{1}{b} - 2$
$= \frac{1}{b} (b^{2} - 2 b + 1) = \frac{1}{b} (b - 1)^{2}$
Required of equation whose roots are
$(α - \frac{1}{β}) and (β - \frac{1}{α}) is,$
$x^{2} - [(α - \frac{1}{β}) + (β - \frac{1}{α})] x + [(α - \frac{1}{β}) \cdot (β - \frac{1}{α})] = 0$
$x^{2} - \frac{a}{b} (1 - b) x + \frac{1}{b} (b - 1)^{2} = 0$
${bx}^{2} + a (b - 1) x + (b - 1)^{2} = 0$

WB JEE-2013

Complex Numbers and Quadratic Equation

118469 The quadratic equation whose sum of the roots is 11 and sum of squares of the roots is 61 is

1

x^{2} + 11 x - 30 = 0

2

x^{2} + 11 x + 30 = 0

3

x^{2} - 11 x - 30 = 0

4

x^{2} - 11 x + 30 = 0

Explanation:

D Sum of the roots $(α + β) = 11$
Square of the $(α^{2} + β^{2}) = 61$
We know that,
$(α + β)^{2} = (α^{2} + β^{2}) + 2 α β$
$(11)^{2} = 61 + 2 α β$
$α β = \frac{121 - 61}{2}$
$α β = 30$
The quadratic equation,
$x^{2} - (α + β) x + (α β) = 0$
$x^{2} - 11 x + 30 = 0$

AP EAMCET-08.07.2022

Complex Numbers and Quadratic Equation

118470 Let $p, q \in R$ . If $2 - \sqrt{3}$ is a root of the quadratic equation, $x^{2} + p + q = 0$ , then

1

q^{2} - 4 p - 16 = 0

2

p^{2} - 4 q - 12 = 0

3

p^{2} - 4 p + 12 = 0

4

q^{2} + 4 p + 14 = 0

Explanation:

B Given,
Let $p, q \in R$ .
If we take root as any real number $β$ , then quadratic equation will be
$x^{2} - (β + 2 - \sqrt{3}) x + β (2 - \sqrt{3}) = 0$
Now, we can have none or any of the options can be correct depending upon ' $β$ '
Instead of $p, q \in R$ it should be $p, q \in R$ it should be $p$ , $q \in Q$ then other root will be $2 + \sqrt{3}$
$\Rightarrow p = - (q + \sqrt{3} - 2 - \sqrt{3}) = - 4$
And, $q = (2 + \sqrt{3}) (2 - \sqrt{3}) = 1$
$\Rightarrow p^{2} - 4 q - 12 = (- 4)^{2} - 4 - 12 = 0$ Hence the correct (b)

JEE Main 09.04.2019

Complex Numbers and Quadratic Equation

118471 The quardratic equation whose roots are $\frac{1}{3 + \sqrt{2}}$ and $\frac{1}{3 - \sqrt{2}}$ , will be

1

7 x^{2} - 6 x + 1 = 0

2

6 x^{2} - 7 x + 1 = 0

3

x^{2} - 6 x + 7 = 0

4

x^{2} - 7 x + 6 = 0

Explanation:

A Given,
The quadratic equation whose roots are
$\frac{1}{3 + \sqrt{2}} and \frac{1}{3 - \sqrt{2}}$
So, the required quadratic equation is $x^{2}$ (sum of the roots) $x +$ product of the roots $= 0$
$x^{2} - (\frac{1}{3 + \sqrt{2}} + \frac{1}{3 - \sqrt{2}}) x + (\frac{1}{3 + \sqrt{2}} \times \frac{1}{3 - \sqrt{2}}) = 0$
$x^{2} - (\frac{3 - \sqrt{2} + 3 + \sqrt{2}}{(3 + \sqrt{2}) (3 - \sqrt{2})}) x + \frac{1}{3^{2} - (\sqrt{2})^{2}} = 0$
$x^{2} - \frac{6}{9 - 2} x + \frac{1}{9 - 2} = 0$
$x^{2} - \frac{6}{7} x + \frac{1}{7} = 0$
$7 x^{2} - 6 x + 1 = 0$

Manipal UGET-2020

Complex Numbers and Quadratic Equation

118472 If $α, β$ are the roots of $a x^{2} + b x + c = 0$ , then the quadratic equation whose roots are $\sqrt{5} α, \sqrt{5} β$ is

1

a x^{2} + \sqrt{5} b x + 5 c = 0

2

a x^{2} + \sqrt{5} bx + \sqrt{5} c = 0

3

a x^{2} + 5 b x + \sqrt{5} c = 0

4

a x^{2} + 5 bx + 5 c = 0

Explanation:

A Given,
If $α, β$ are the roots of
$a x^{2} + b x + c = 0$
$α + β = - \frac{b}{a}$
$α \cdot β = \frac{c}{a}$
If the roots of quadratic equation is $\sqrt{5} α$ and $\sqrt{5} β$ then the equation.
$x^{2} - (Sum of roots) x + (product of roots) = 0$
$\Rightarrow α = \frac{- b}{a} β = \frac{c}{a}$
$\Rightarrow x^{2} - (\sqrt{5} α + \sqrt{5} β) x + (\sqrt{5} α \times \sqrt{5} β) = 0$
$\Rightarrow x^{2} - \sqrt{5} (α + β) x + 5 α β = 0$
$\Rightarrow x^{2} - \sqrt{5} (\frac{- b}{a}) x + \frac{5 c}{a} = 0$
$\Rightarrow {ax}^{2} + \sqrt{5} b x + 5 c = 0$

AP EAMCET-07.07.2022

Complex Numbers and Quadratic Equation

118468 If $α, β$ are the roots of the quadratic equation $x^{2} + a x + b = 0, (b \neq 0)$ , then the quadratic equation whose roots are $α - \frac{1}{β}, β - \frac{1}{α}$ , is

1

a x^{2} + a (b - 1) x + (a - 1)^{2} = 0

2

b x^{2} + a (b - 1) x + (b - 1)^{2} = 0

3

x^{2} + a x + b v = 0

4

a b x^{2} + b x + a = 0

Explanation:

B Given,
$x^{2} + a x + b = 0 b \neq 0$
$α + β = - a$
$α \cdot β = b$
Now,
$(α - \frac{1}{β}) + (β - \frac{1}{α}) = (α + β) - (\frac{α + β}{α \cdot β})$
$= (- a) - (\frac{- a}{b})$
$= - a + \frac{a}{b}$
$= - \frac{a}{b} (1 - b)$
And,
$(α - \frac{1}{β}) (β - \frac{1}{α}) = α β - 1 - 1 + \frac{1}{α β}$
$= b + \frac{1}{b} - 2$
$= \frac{1}{b} (b^{2} - 2 b + 1) = \frac{1}{b} (b - 1)^{2}$
Required of equation whose roots are
$(α - \frac{1}{β}) and (β - \frac{1}{α}) is,$
$x^{2} - [(α - \frac{1}{β}) + (β - \frac{1}{α})] x + [(α - \frac{1}{β}) \cdot (β - \frac{1}{α})] = 0$
$x^{2} - \frac{a}{b} (1 - b) x + \frac{1}{b} (b - 1)^{2} = 0$
${bx}^{2} + a (b - 1) x + (b - 1)^{2} = 0$

WB JEE-2013

Complex Numbers and Quadratic Equation

118469 The quadratic equation whose sum of the roots is 11 and sum of squares of the roots is 61 is

1

x^{2} + 11 x - 30 = 0

2

x^{2} + 11 x + 30 = 0

3

x^{2} - 11 x - 30 = 0

4

x^{2} - 11 x + 30 = 0

Explanation:

D Sum of the roots $(α + β) = 11$
Square of the $(α^{2} + β^{2}) = 61$
We know that,
$(α + β)^{2} = (α^{2} + β^{2}) + 2 α β$
$(11)^{2} = 61 + 2 α β$
$α β = \frac{121 - 61}{2}$
$α β = 30$
The quadratic equation,
$x^{2} - (α + β) x + (α β) = 0$
$x^{2} - 11 x + 30 = 0$

AP EAMCET-08.07.2022

Complex Numbers and Quadratic Equation

118470 Let $p, q \in R$ . If $2 - \sqrt{3}$ is a root of the quadratic equation, $x^{2} + p + q = 0$ , then

1

q^{2} - 4 p - 16 = 0

2

p^{2} - 4 q - 12 = 0

3

p^{2} - 4 p + 12 = 0

4

q^{2} + 4 p + 14 = 0

Explanation:

B Given,
Let $p, q \in R$ .
If we take root as any real number $β$ , then quadratic equation will be
$x^{2} - (β + 2 - \sqrt{3}) x + β (2 - \sqrt{3}) = 0$
Now, we can have none or any of the options can be correct depending upon ' $β$ '
Instead of $p, q \in R$ it should be $p, q \in R$ it should be $p$ , $q \in Q$ then other root will be $2 + \sqrt{3}$
$\Rightarrow p = - (q + \sqrt{3} - 2 - \sqrt{3}) = - 4$
And, $q = (2 + \sqrt{3}) (2 - \sqrt{3}) = 1$
$\Rightarrow p^{2} - 4 q - 12 = (- 4)^{2} - 4 - 12 = 0$ Hence the correct (b)

JEE Main 09.04.2019

Complex Numbers and Quadratic Equation

118471 The quardratic equation whose roots are $\frac{1}{3 + \sqrt{2}}$ and $\frac{1}{3 - \sqrt{2}}$ , will be

1

7 x^{2} - 6 x + 1 = 0

2

6 x^{2} - 7 x + 1 = 0

3

x^{2} - 6 x + 7 = 0

4

x^{2} - 7 x + 6 = 0

Explanation:

A Given,
The quadratic equation whose roots are
$\frac{1}{3 + \sqrt{2}} and \frac{1}{3 - \sqrt{2}}$
So, the required quadratic equation is $x^{2}$ (sum of the roots) $x +$ product of the roots $= 0$
$x^{2} - (\frac{1}{3 + \sqrt{2}} + \frac{1}{3 - \sqrt{2}}) x + (\frac{1}{3 + \sqrt{2}} \times \frac{1}{3 - \sqrt{2}}) = 0$
$x^{2} - (\frac{3 - \sqrt{2} + 3 + \sqrt{2}}{(3 + \sqrt{2}) (3 - \sqrt{2})}) x + \frac{1}{3^{2} - (\sqrt{2})^{2}} = 0$
$x^{2} - \frac{6}{9 - 2} x + \frac{1}{9 - 2} = 0$
$x^{2} - \frac{6}{7} x + \frac{1}{7} = 0$
$7 x^{2} - 6 x + 1 = 0$

Manipal UGET-2020

Complex Numbers and Quadratic Equation

118472 If $α, β$ are the roots of $a x^{2} + b x + c = 0$ , then the quadratic equation whose roots are $\sqrt{5} α, \sqrt{5} β$ is

1

a x^{2} + \sqrt{5} b x + 5 c = 0

2

a x^{2} + \sqrt{5} bx + \sqrt{5} c = 0

3

a x^{2} + 5 b x + \sqrt{5} c = 0

4

a x^{2} + 5 bx + 5 c = 0

Explanation:

A Given,
If $α, β$ are the roots of
$a x^{2} + b x + c = 0$
$α + β = - \frac{b}{a}$
$α \cdot β = \frac{c}{a}$
If the roots of quadratic equation is $\sqrt{5} α$ and $\sqrt{5} β$ then the equation.
$x^{2} - (Sum of roots) x + (product of roots) = 0$
$\Rightarrow α = \frac{- b}{a} β = \frac{c}{a}$
$\Rightarrow x^{2} - (\sqrt{5} α + \sqrt{5} β) x + (\sqrt{5} α \times \sqrt{5} β) = 0$
$\Rightarrow x^{2} - \sqrt{5} (α + β) x + 5 α β = 0$
$\Rightarrow x^{2} - \sqrt{5} (\frac{- b}{a}) x + \frac{5 c}{a} = 0$
$\Rightarrow {ax}^{2} + \sqrt{5} b x + 5 c = 0$

AP EAMCET-07.07.2022