QBManager

Complex Numbers and Quadratic Equation

118279 If $α, β$ are the real roots of $x^{2} + p x + q = 0$ and $α^{4}, β^{4}$ are the roots of $x^{2} - r x + s = 0$ , then the equation $x^{2} - 4 q x + 2 q^{2} - r = 0$ has always

1 two positive roots

2 two negative roots

3 one positive root and one negative root

4 two imaginary roots

Explanation:

C Given, $α, β$ are the real roots of $x^{2} + p x + q = 0 = 0$
$α + β = - p$
$α \cdot β = q$
We know that,
$α^{2} + β^{2} = (α + β)^{2} - 2 α β$
$= p^{2} - 2 q$
$α^{4} + β^{4} = {(α^{2} + β^{2})}^{2} - 2 α^{2} β^{2}$
$= {(p^{2} - 2 q)}^{2} - 2 q^{2}$
$α^{4}, β^{4}$ are the roots of $x^{2} - r x + s = 0$
$α^{4} + β^{4} = r$
$α^{4} \times β^{4} = s$
$α^{4} + β^{4} = {(p^{2} - 2 q)}^{2} - 2 q^{2}$
$r = {(p^{2} - 2 q)}^{2} - 2 q^{2}$
$r + 2 q^{2} = {(p^{2} - 2 q)}^{2}$
The roots of the equation $x^{2} - 4 q x + 2 q^{2} - r = 0$
$x = \frac{4 q \pm \sqrt{(4 q)^{2} - 4 (2 q^{2} - r)}}{2}$
$x = 2 q \pm \sqrt{2 q^{2} + r}$
From equation (i),
$x = 2 q \pm \sqrt{{(p^{2} - 2 q)}^{2}}$
$x = 2 q \pm (p^{2} - 2 q)$
$One of the root = 2 q + p^{2} - 2 q$
$= p^{2}, which is positive$
$Second root is = 2 q - p^{2} + 2 q$
$= 4 q - p^{2}$
$∴ p^{2} - 4 q > 0, since the first quadratic has two real$
$roots$
$4 q - p^{2} is negative.$
Second root is $= 2 q - p^{2} + 2 q$
$∴ p^{2} - 4 q > 0$ , since the first quadratic has two real roots $4 q - p^{2}$ is negative.

AP EAMCET-21.04.2019

Complex Numbers and Quadratic Equation

118280 If $α, β$ are the roots of the equation $x^{2} + 2 x + 4$ $= 0$ , then $\frac{1}{α^{2}} + \frac{1}{β^{2}}$ equal to

1

\frac{- 1}{4}

2

\frac{1}{4}

3 32

4

\frac{1}{32}

Explanation:

A Given, $α, β$ are the roots of the equations $x^{2} + 2 x + 4 = 0$
$α + β = - 2$
$α \cdot β = 4$
$\frac{1}{α^{2}} + \frac{1}{β^{2}} = \frac{(α + β)^{2} - 2 α \cdot β}{(α \cdot β)^{2}}$
$= \frac{(- 2)^{2} - 2 (4)}{(4)^{2}} = \frac{4 - 8}{16} = \frac{- 1}{4}$

AP EAMCET-05.10.2021

Complex Numbers and Quadratic Equation

118282 If $α$ and $β$ are the roots of the quadratic equation $a x^{2} + b x + c = 0$ and $3 b^{2} = 16 a c$ , then

1

α = 4 β

or

β = 4 α

2

α = - 4 β

or

β = 4 α

3

α = 3 β

or

β = 3 α

4

α = - 3 β

or

β = - 3 α

Explanation:

C Given,
$a x^{2} + b x + c = 0$
Sum of roots, $α + β = \frac{- b}{a}$
Product of roots $α \cdot β = \frac{c}{a}$
Given,
$3 b^{2} = 16 a c$
$3 a^{2} (α + β)^{2} = 16 a \cdot a (α β)$
$3 α^{2} + 3 β^{2} + 6 α β = 16 α β$
$3 α^{2} - 10 α β + 3 β^{2} = 0$
$(α - 3 β) (3 α - β) = 0$
$α = 3 β, β = 3 α$

WB JEE-2013

Complex Numbers and Quadratic Equation

118283 If one root of $x^{2} - x - k = 0$ is square of the other, then $k$ is equal to

1

2 \pm \sqrt{3}

2

3 \pm \sqrt{2}

3

2 \pm \sqrt{5}

4

5 \pm \sqrt{2}

Explanation:

C Given that one root of $x^{2} - x - k = 0$ is square of the other then we have to find $k$ is = ?
Let us consider the roots be $α, α^{2}$
The quadratic equation is
$(x - α) (x - α^{2}) = 0$
$x^{2} - (α + α^{2}) x + α^{3} = 0$
Comparing this with the given quadratic equation
$x^{2} - x - k = x^{2} - (α + α^{2}) x + α^{3}$
Comparing the coefficient we get
$α + α^{2} = 1 and k = - α^{3}$
$α + α^{2} = 1$
${(α + α^{2})}^{3} = 1$
$α^{3} + α^{6} + 3 α^{3} (α + α^{2}) = 1$
$k = - α^{3}$
Substituting this in equation (i)
$- k + k^{2} + 3 (- k) (1) = 1$
$k^{2} - 4 k - 1 = 0$
$k = \frac{4 \pm \sqrt{16 + 4}}{2}$
$k = 2 \pm \sqrt{5}$

Jamia Millia Islamia-2010

Complex Numbers and Quadratic Equation

118279 If $α, β$ are the real roots of $x^{2} + p x + q = 0$ and $α^{4}, β^{4}$ are the roots of $x^{2} - r x + s = 0$ , then the equation $x^{2} - 4 q x + 2 q^{2} - r = 0$ has always

1 two positive roots

2 two negative roots

3 one positive root and one negative root

4 two imaginary roots

Explanation:

C Given, $α, β$ are the real roots of $x^{2} + p x + q = 0 = 0$
$α + β = - p$
$α \cdot β = q$
We know that,
$α^{2} + β^{2} = (α + β)^{2} - 2 α β$
$= p^{2} - 2 q$
$α^{4} + β^{4} = {(α^{2} + β^{2})}^{2} - 2 α^{2} β^{2}$
$= {(p^{2} - 2 q)}^{2} - 2 q^{2}$
$α^{4}, β^{4}$ are the roots of $x^{2} - r x + s = 0$
$α^{4} + β^{4} = r$
$α^{4} \times β^{4} = s$
$α^{4} + β^{4} = {(p^{2} - 2 q)}^{2} - 2 q^{2}$
$r = {(p^{2} - 2 q)}^{2} - 2 q^{2}$
$r + 2 q^{2} = {(p^{2} - 2 q)}^{2}$
The roots of the equation $x^{2} - 4 q x + 2 q^{2} - r = 0$
$x = \frac{4 q \pm \sqrt{(4 q)^{2} - 4 (2 q^{2} - r)}}{2}$
$x = 2 q \pm \sqrt{2 q^{2} + r}$
From equation (i),
$x = 2 q \pm \sqrt{{(p^{2} - 2 q)}^{2}}$
$x = 2 q \pm (p^{2} - 2 q)$
$One of the root = 2 q + p^{2} - 2 q$
$= p^{2}, which is positive$
$Second root is = 2 q - p^{2} + 2 q$
$= 4 q - p^{2}$
$∴ p^{2} - 4 q > 0, since the first quadratic has two real$
$roots$
$4 q - p^{2} is negative.$
Second root is $= 2 q - p^{2} + 2 q$
$∴ p^{2} - 4 q > 0$ , since the first quadratic has two real roots $4 q - p^{2}$ is negative.

AP EAMCET-21.04.2019

Complex Numbers and Quadratic Equation

118280 If $α, β$ are the roots of the equation $x^{2} + 2 x + 4$ $= 0$ , then $\frac{1}{α^{2}} + \frac{1}{β^{2}}$ equal to

1

\frac{- 1}{4}

2

\frac{1}{4}

3 32

4

\frac{1}{32}

Explanation:

A Given, $α, β$ are the roots of the equations $x^{2} + 2 x + 4 = 0$
$α + β = - 2$
$α \cdot β = 4$
$\frac{1}{α^{2}} + \frac{1}{β^{2}} = \frac{(α + β)^{2} - 2 α \cdot β}{(α \cdot β)^{2}}$
$= \frac{(- 2)^{2} - 2 (4)}{(4)^{2}} = \frac{4 - 8}{16} = \frac{- 1}{4}$

AP EAMCET-05.10.2021

Complex Numbers and Quadratic Equation

118282 If $α$ and $β$ are the roots of the quadratic equation $a x^{2} + b x + c = 0$ and $3 b^{2} = 16 a c$ , then

1

α = 4 β

or

β = 4 α

2

α = - 4 β

or

β = 4 α

3

α = 3 β

or

β = 3 α

4

α = - 3 β

or

β = - 3 α

Explanation:

C Given,
$a x^{2} + b x + c = 0$
Sum of roots, $α + β = \frac{- b}{a}$
Product of roots $α \cdot β = \frac{c}{a}$
Given,
$3 b^{2} = 16 a c$
$3 a^{2} (α + β)^{2} = 16 a \cdot a (α β)$
$3 α^{2} + 3 β^{2} + 6 α β = 16 α β$
$3 α^{2} - 10 α β + 3 β^{2} = 0$
$(α - 3 β) (3 α - β) = 0$
$α = 3 β, β = 3 α$

WB JEE-2013

Complex Numbers and Quadratic Equation

118283 If one root of $x^{2} - x - k = 0$ is square of the other, then $k$ is equal to

1

2 \pm \sqrt{3}

2

3 \pm \sqrt{2}

3

2 \pm \sqrt{5}

4

5 \pm \sqrt{2}

Explanation:

C Given that one root of $x^{2} - x - k = 0$ is square of the other then we have to find $k$ is = ?
Let us consider the roots be $α, α^{2}$
The quadratic equation is
$(x - α) (x - α^{2}) = 0$
$x^{2} - (α + α^{2}) x + α^{3} = 0$
Comparing this with the given quadratic equation
$x^{2} - x - k = x^{2} - (α + α^{2}) x + α^{3}$
Comparing the coefficient we get
$α + α^{2} = 1 and k = - α^{3}$
$α + α^{2} = 1$
${(α + α^{2})}^{3} = 1$
$α^{3} + α^{6} + 3 α^{3} (α + α^{2}) = 1$
$k = - α^{3}$
Substituting this in equation (i)
$- k + k^{2} + 3 (- k) (1) = 1$
$k^{2} - 4 k - 1 = 0$
$k = \frac{4 \pm \sqrt{16 + 4}}{2}$
$k = 2 \pm \sqrt{5}$

Jamia Millia Islamia-2010

Complex Numbers and Quadratic Equation

118279 If $α, β$ are the real roots of $x^{2} + p x + q = 0$ and $α^{4}, β^{4}$ are the roots of $x^{2} - r x + s = 0$ , then the equation $x^{2} - 4 q x + 2 q^{2} - r = 0$ has always

1 two positive roots

2 two negative roots

3 one positive root and one negative root

4 two imaginary roots

Explanation:

C Given, $α, β$ are the real roots of $x^{2} + p x + q = 0 = 0$
$α + β = - p$
$α \cdot β = q$
We know that,
$α^{2} + β^{2} = (α + β)^{2} - 2 α β$
$= p^{2} - 2 q$
$α^{4} + β^{4} = {(α^{2} + β^{2})}^{2} - 2 α^{2} β^{2}$
$= {(p^{2} - 2 q)}^{2} - 2 q^{2}$
$α^{4}, β^{4}$ are the roots of $x^{2} - r x + s = 0$
$α^{4} + β^{4} = r$
$α^{4} \times β^{4} = s$
$α^{4} + β^{4} = {(p^{2} - 2 q)}^{2} - 2 q^{2}$
$r = {(p^{2} - 2 q)}^{2} - 2 q^{2}$
$r + 2 q^{2} = {(p^{2} - 2 q)}^{2}$
The roots of the equation $x^{2} - 4 q x + 2 q^{2} - r = 0$
$x = \frac{4 q \pm \sqrt{(4 q)^{2} - 4 (2 q^{2} - r)}}{2}$
$x = 2 q \pm \sqrt{2 q^{2} + r}$
From equation (i),
$x = 2 q \pm \sqrt{{(p^{2} - 2 q)}^{2}}$
$x = 2 q \pm (p^{2} - 2 q)$
$One of the root = 2 q + p^{2} - 2 q$
$= p^{2}, which is positive$
$Second root is = 2 q - p^{2} + 2 q$
$= 4 q - p^{2}$
$∴ p^{2} - 4 q > 0, since the first quadratic has two real$
$roots$
$4 q - p^{2} is negative.$
Second root is $= 2 q - p^{2} + 2 q$
$∴ p^{2} - 4 q > 0$ , since the first quadratic has two real roots $4 q - p^{2}$ is negative.

AP EAMCET-21.04.2019

Complex Numbers and Quadratic Equation

118280 If $α, β$ are the roots of the equation $x^{2} + 2 x + 4$ $= 0$ , then $\frac{1}{α^{2}} + \frac{1}{β^{2}}$ equal to

1

\frac{- 1}{4}

2

\frac{1}{4}

3 32

4

\frac{1}{32}

Explanation:

A Given, $α, β$ are the roots of the equations $x^{2} + 2 x + 4 = 0$
$α + β = - 2$
$α \cdot β = 4$
$\frac{1}{α^{2}} + \frac{1}{β^{2}} = \frac{(α + β)^{2} - 2 α \cdot β}{(α \cdot β)^{2}}$
$= \frac{(- 2)^{2} - 2 (4)}{(4)^{2}} = \frac{4 - 8}{16} = \frac{- 1}{4}$

AP EAMCET-05.10.2021

Complex Numbers and Quadratic Equation

118282 If $α$ and $β$ are the roots of the quadratic equation $a x^{2} + b x + c = 0$ and $3 b^{2} = 16 a c$ , then

1

α = 4 β

or

β = 4 α

2

α = - 4 β

or

β = 4 α

3

α = 3 β

or

β = 3 α

4

α = - 3 β

or

β = - 3 α

Explanation:

C Given,
$a x^{2} + b x + c = 0$
Sum of roots, $α + β = \frac{- b}{a}$
Product of roots $α \cdot β = \frac{c}{a}$
Given,
$3 b^{2} = 16 a c$
$3 a^{2} (α + β)^{2} = 16 a \cdot a (α β)$
$3 α^{2} + 3 β^{2} + 6 α β = 16 α β$
$3 α^{2} - 10 α β + 3 β^{2} = 0$
$(α - 3 β) (3 α - β) = 0$
$α = 3 β, β = 3 α$

WB JEE-2013

Complex Numbers and Quadratic Equation

118283 If one root of $x^{2} - x - k = 0$ is square of the other, then $k$ is equal to

1

2 \pm \sqrt{3}

2

3 \pm \sqrt{2}

3

2 \pm \sqrt{5}

4

5 \pm \sqrt{2}

Explanation:

C Given that one root of $x^{2} - x - k = 0$ is square of the other then we have to find $k$ is = ?
Let us consider the roots be $α, α^{2}$
The quadratic equation is
$(x - α) (x - α^{2}) = 0$
$x^{2} - (α + α^{2}) x + α^{3} = 0$
Comparing this with the given quadratic equation
$x^{2} - x - k = x^{2} - (α + α^{2}) x + α^{3}$
Comparing the coefficient we get
$α + α^{2} = 1 and k = - α^{3}$
$α + α^{2} = 1$
${(α + α^{2})}^{3} = 1$
$α^{3} + α^{6} + 3 α^{3} (α + α^{2}) = 1$
$k = - α^{3}$
Substituting this in equation (i)
$- k + k^{2} + 3 (- k) (1) = 1$
$k^{2} - 4 k - 1 = 0$
$k = \frac{4 \pm \sqrt{16 + 4}}{2}$
$k = 2 \pm \sqrt{5}$

Jamia Millia Islamia-2010

Complex Numbers and Quadratic Equation

118279 If $α, β$ are the real roots of $x^{2} + p x + q = 0$ and $α^{4}, β^{4}$ are the roots of $x^{2} - r x + s = 0$ , then the equation $x^{2} - 4 q x + 2 q^{2} - r = 0$ has always

1 two positive roots

2 two negative roots

3 one positive root and one negative root

4 two imaginary roots

Explanation:

C Given, $α, β$ are the real roots of $x^{2} + p x + q = 0 = 0$
$α + β = - p$
$α \cdot β = q$
We know that,
$α^{2} + β^{2} = (α + β)^{2} - 2 α β$
$= p^{2} - 2 q$
$α^{4} + β^{4} = {(α^{2} + β^{2})}^{2} - 2 α^{2} β^{2}$
$= {(p^{2} - 2 q)}^{2} - 2 q^{2}$
$α^{4}, β^{4}$ are the roots of $x^{2} - r x + s = 0$
$α^{4} + β^{4} = r$
$α^{4} \times β^{4} = s$
$α^{4} + β^{4} = {(p^{2} - 2 q)}^{2} - 2 q^{2}$
$r = {(p^{2} - 2 q)}^{2} - 2 q^{2}$
$r + 2 q^{2} = {(p^{2} - 2 q)}^{2}$
The roots of the equation $x^{2} - 4 q x + 2 q^{2} - r = 0$
$x = \frac{4 q \pm \sqrt{(4 q)^{2} - 4 (2 q^{2} - r)}}{2}$
$x = 2 q \pm \sqrt{2 q^{2} + r}$
From equation (i),
$x = 2 q \pm \sqrt{{(p^{2} - 2 q)}^{2}}$
$x = 2 q \pm (p^{2} - 2 q)$
$One of the root = 2 q + p^{2} - 2 q$
$= p^{2}, which is positive$
$Second root is = 2 q - p^{2} + 2 q$
$= 4 q - p^{2}$
$∴ p^{2} - 4 q > 0, since the first quadratic has two real$
$roots$
$4 q - p^{2} is negative.$
Second root is $= 2 q - p^{2} + 2 q$
$∴ p^{2} - 4 q > 0$ , since the first quadratic has two real roots $4 q - p^{2}$ is negative.

AP EAMCET-21.04.2019

Complex Numbers and Quadratic Equation

118280 If $α, β$ are the roots of the equation $x^{2} + 2 x + 4$ $= 0$ , then $\frac{1}{α^{2}} + \frac{1}{β^{2}}$ equal to

1

\frac{- 1}{4}

2

\frac{1}{4}

3 32

4

\frac{1}{32}

Explanation:

A Given, $α, β$ are the roots of the equations $x^{2} + 2 x + 4 = 0$
$α + β = - 2$
$α \cdot β = 4$
$\frac{1}{α^{2}} + \frac{1}{β^{2}} = \frac{(α + β)^{2} - 2 α \cdot β}{(α \cdot β)^{2}}$
$= \frac{(- 2)^{2} - 2 (4)}{(4)^{2}} = \frac{4 - 8}{16} = \frac{- 1}{4}$

AP EAMCET-05.10.2021

Complex Numbers and Quadratic Equation

118282 If $α$ and $β$ are the roots of the quadratic equation $a x^{2} + b x + c = 0$ and $3 b^{2} = 16 a c$ , then

1

α = 4 β

or

β = 4 α

2

α = - 4 β

or

β = 4 α

3

α = 3 β

or

β = 3 α

4

α = - 3 β

or

β = - 3 α

Explanation:

C Given,
$a x^{2} + b x + c = 0$
Sum of roots, $α + β = \frac{- b}{a}$
Product of roots $α \cdot β = \frac{c}{a}$
Given,
$3 b^{2} = 16 a c$
$3 a^{2} (α + β)^{2} = 16 a \cdot a (α β)$
$3 α^{2} + 3 β^{2} + 6 α β = 16 α β$
$3 α^{2} - 10 α β + 3 β^{2} = 0$
$(α - 3 β) (3 α - β) = 0$
$α = 3 β, β = 3 α$

WB JEE-2013

Complex Numbers and Quadratic Equation

118283 If one root of $x^{2} - x - k = 0$ is square of the other, then $k$ is equal to

1

2 \pm \sqrt{3}

2

3 \pm \sqrt{2}

3

2 \pm \sqrt{5}

4

5 \pm \sqrt{2}

Explanation:

C Given that one root of $x^{2} - x - k = 0$ is square of the other then we have to find $k$ is = ?
Let us consider the roots be $α, α^{2}$
The quadratic equation is
$(x - α) (x - α^{2}) = 0$
$x^{2} - (α + α^{2}) x + α^{3} = 0$
Comparing this with the given quadratic equation
$x^{2} - x - k = x^{2} - (α + α^{2}) x + α^{3}$
Comparing the coefficient we get
$α + α^{2} = 1 and k = - α^{3}$
$α + α^{2} = 1$
${(α + α^{2})}^{3} = 1$
$α^{3} + α^{6} + 3 α^{3} (α + α^{2}) = 1$
$k = - α^{3}$
Substituting this in equation (i)
$- k + k^{2} + 3 (- k) (1) = 1$
$k^{2} - 4 k - 1 = 0$
$k = \frac{4 \pm \sqrt{16 + 4}}{2}$
$k = 2 \pm \sqrt{5}$

Jamia Millia Islamia-2010