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Complex Numbers and Quadratic Equation

118473 The equation of lowest degree with rational coefficients having roots $\sqrt{3} + \sqrt{2} i$ and $\sqrt{3} - \sqrt{2}$ is

1

(x^{4} - 2 x^{2} + 25) (x^{4} - 10 x^{2} + 1) = 0

2

(x^{2} - 2 \sqrt{3} x + 5) (x^{2} - 2 \sqrt{3} x + 1) = 0

3

(x^{4} - 2 x^{2} + 25) (x^{4} + 10 x^{2} + 1) = 0

4

(x^{4} - 10 x^{2} + 1) (x^{4} + 2 x^{2} + 25) = 0

Explanation:

A The roots of the given equations are
$\sqrt{3} + \sqrt{2} i and \sqrt{3} - \sqrt{2}$
$x = \sqrt{3} + \sqrt{2} i$
$x - \sqrt{3} = \sqrt{2} i$
On squaring both sides
$x^{2} + 3 - 2 \sqrt{3} x = 2 i^{2}$
$x^{2} + 3 - 2 \sqrt{3} x = - 2$
$x^{2} + 5 = 2 \sqrt{3} x$
Again on squaring both sides
$x^{4} + 25 + 10 x^{2} = 12 x^{2}$
$x^{4} - 2 x^{2} + 25 = 0$
Also, $x = \sqrt{3} - \sqrt{2}$
On squaring both sides
$x^{2} = (\sqrt{3} - \sqrt{2})^{2}$
$x^{2} = 3 + 2 - 2 \sqrt{6}$
$x^{2} - 5 = - 2 \sqrt{6}$
Again on squaring both sides
${(x^{2} - 5)}^{2} = (- 2 \sqrt{6})^{2}$
$x^{4} + 25 - 10 x^{2} = 24$
$x^{4} - 10 x^{2} + 1 = 0$
$∴$ The required equation is
$(x^{4} - 2 x^{2} + 25) (x^{4} - 10 x^{2} + 1) = 0$

TS EAMCET-19.07.2022

Complex Numbers and Quadratic Equation

118474 $p$ is non-zero real number. If the equation whose roots are the squares of the roots of the equation $x^{3} - p x^{2} + p x - 1 = 0$ is identical with the given equation, then $p =$

1

\frac{1}{2}

2 2

3 3

4 -1

Explanation:

C Let $α, β, γ$ are roots of equation
$x^{3} - {px}^{2} + px - 1 = 0$
$α + β + γ = p$
$α β + β γ + α γ = p$
$α β γ = 1$
Also, given,
$α^{2} + β^{2} + γ^{2} = p$
$α^{2} β^{2} + β^{2} γ^{2} + α^{2} γ^{2} = p$
$α^{2} β^{2} γ^{2} = 1$
$∴ (α + β + γ)^{2} = α^{2} + β^{2} + γ^{2} + 2 (α β + β γ + γ α)$
$p^{2} = p + 2 p$
$p^{2} = 3 p$
$p = 3$

TS EAMCET-10.09.2020

Complex Numbers and Quadratic Equation

118475 The equation whose roots are squares of the roots of $x^{4} - 2 x^{3} + 6 x - 21 = 0$ is

1

x^{4} - 4 x^{3} - 18 x^{2} - 36 x + 441 = 0

2

x^{4} + 18 x^{3} - 4 x^{2} + 36 x + 441 = 0

3

x^{4} - 2 x^{3} + 4 x^{2} + 6 x + 441 = 0

4

x^{4} + 3 x^{3} - 5 x^{2} + 6 x + 441 = 0

Explanation:

Exp: (A): Let, $f (x) = x^{4} - 2 x^{3} + 6 x - 21 = 0$
Required equation where roots are squares of the roots of $f (x)$ is.
$f (\sqrt{x}) = 0$
$(\sqrt{x})^{4} - 2 (\sqrt{x})^{3} + 6 \sqrt{x} - 21 = 0$
$\Rightarrow x^{2} - 2 x \sqrt{x} + 6 \sqrt{x} - 21 = 0$
$\Rightarrow x^{2} - 21 = 2 x \sqrt{x} - 6 \sqrt{x}$
$\Rightarrow x^{2} - 21 = 2 \sqrt{x} (x - 3)$
$\Rightarrow x^{2} - 21 = 2 \sqrt{x} (x - 3)$
$\Rightarrow {(x^{2} - 21)}^{2} = 4 x (x - 3)^{2}$
$\Rightarrow x^{4} - 42 x^{2} + 441 = 4 x (x^{2} - 6 x + 9)$
$\Rightarrow x^{4} - 42 x^{2} + 441 = 4 x^{3} - 24 x^{2} + 36 x$
$\Rightarrow x^{4} - 4 x^{3} - 18 x^{2} - 36 x + 441 = 0$

TS EAMCET-04.08.2021

Complex Numbers and Quadratic Equation

118476 If $\frac{α}{α + 1}$ and $\frac{β}{β + 1}$ are the roots of the quadratic equation $x^{2} + 7 x + 3 = 0$ , then the equation having roots $α$ and $β$ is

1

3 x^{2} - x - 3 = 0

2

11 x^{2} + 13 x + 3 = 0

3

13 x^{2} + 11 x + 13 = 0

4

11 x^{2} + 3 x + 13 = 0

Explanation:

B Let, $f (x) = x^{2} + 7 x + 3 = 0$
Roots are $\frac{α}{α + 1}$ and $\frac{β}{β + 1}$
Equation having roots $α$ and $β$
$Is, f (\frac{x}{x + 1}) = 0$
${(\frac{x}{x + 1})}^{2} + 7 (\frac{x}{x + 1}) + 3 = 0$
$x^{2} + 7 x (x + 1) + 3 (x + 1)^{2} = 0$
$x^{2} + 7 x^{2} + 7 x + 3 (x^{2} + 1 + 2 x) = 0$
$x^{2} + 7 x^{2} + 7 x + 3 x^{2} + 3 + 6 x = 0$
$11 x^{2} + 13 x + 3 = 0$

TS EAMCET-04.08.2021

Complex Numbers and Quadratic Equation

118473 The equation of lowest degree with rational coefficients having roots $\sqrt{3} + \sqrt{2} i$ and $\sqrt{3} - \sqrt{2}$ is

1

(x^{4} - 2 x^{2} + 25) (x^{4} - 10 x^{2} + 1) = 0

2

(x^{2} - 2 \sqrt{3} x + 5) (x^{2} - 2 \sqrt{3} x + 1) = 0

3

(x^{4} - 2 x^{2} + 25) (x^{4} + 10 x^{2} + 1) = 0

4

(x^{4} - 10 x^{2} + 1) (x^{4} + 2 x^{2} + 25) = 0

Explanation:

A The roots of the given equations are
$\sqrt{3} + \sqrt{2} i and \sqrt{3} - \sqrt{2}$
$x = \sqrt{3} + \sqrt{2} i$
$x - \sqrt{3} = \sqrt{2} i$
On squaring both sides
$x^{2} + 3 - 2 \sqrt{3} x = 2 i^{2}$
$x^{2} + 3 - 2 \sqrt{3} x = - 2$
$x^{2} + 5 = 2 \sqrt{3} x$
Again on squaring both sides
$x^{4} + 25 + 10 x^{2} = 12 x^{2}$
$x^{4} - 2 x^{2} + 25 = 0$
Also, $x = \sqrt{3} - \sqrt{2}$
On squaring both sides
$x^{2} = (\sqrt{3} - \sqrt{2})^{2}$
$x^{2} = 3 + 2 - 2 \sqrt{6}$
$x^{2} - 5 = - 2 \sqrt{6}$
Again on squaring both sides
${(x^{2} - 5)}^{2} = (- 2 \sqrt{6})^{2}$
$x^{4} + 25 - 10 x^{2} = 24$
$x^{4} - 10 x^{2} + 1 = 0$
$∴$ The required equation is
$(x^{4} - 2 x^{2} + 25) (x^{4} - 10 x^{2} + 1) = 0$

TS EAMCET-19.07.2022

Complex Numbers and Quadratic Equation

118474 $p$ is non-zero real number. If the equation whose roots are the squares of the roots of the equation $x^{3} - p x^{2} + p x - 1 = 0$ is identical with the given equation, then $p =$

1

\frac{1}{2}

2 2

3 3

4 -1

Explanation:

C Let $α, β, γ$ are roots of equation
$x^{3} - {px}^{2} + px - 1 = 0$
$α + β + γ = p$
$α β + β γ + α γ = p$
$α β γ = 1$
Also, given,
$α^{2} + β^{2} + γ^{2} = p$
$α^{2} β^{2} + β^{2} γ^{2} + α^{2} γ^{2} = p$
$α^{2} β^{2} γ^{2} = 1$
$∴ (α + β + γ)^{2} = α^{2} + β^{2} + γ^{2} + 2 (α β + β γ + γ α)$
$p^{2} = p + 2 p$
$p^{2} = 3 p$
$p = 3$

TS EAMCET-10.09.2020

Complex Numbers and Quadratic Equation

118475 The equation whose roots are squares of the roots of $x^{4} - 2 x^{3} + 6 x - 21 = 0$ is

1

x^{4} - 4 x^{3} - 18 x^{2} - 36 x + 441 = 0

2

x^{4} + 18 x^{3} - 4 x^{2} + 36 x + 441 = 0

3

x^{4} - 2 x^{3} + 4 x^{2} + 6 x + 441 = 0

4

x^{4} + 3 x^{3} - 5 x^{2} + 6 x + 441 = 0

Explanation:

Exp: (A): Let, $f (x) = x^{4} - 2 x^{3} + 6 x - 21 = 0$
Required equation where roots are squares of the roots of $f (x)$ is.
$f (\sqrt{x}) = 0$
$(\sqrt{x})^{4} - 2 (\sqrt{x})^{3} + 6 \sqrt{x} - 21 = 0$
$\Rightarrow x^{2} - 2 x \sqrt{x} + 6 \sqrt{x} - 21 = 0$
$\Rightarrow x^{2} - 21 = 2 x \sqrt{x} - 6 \sqrt{x}$
$\Rightarrow x^{2} - 21 = 2 \sqrt{x} (x - 3)$
$\Rightarrow x^{2} - 21 = 2 \sqrt{x} (x - 3)$
$\Rightarrow {(x^{2} - 21)}^{2} = 4 x (x - 3)^{2}$
$\Rightarrow x^{4} - 42 x^{2} + 441 = 4 x (x^{2} - 6 x + 9)$
$\Rightarrow x^{4} - 42 x^{2} + 441 = 4 x^{3} - 24 x^{2} + 36 x$
$\Rightarrow x^{4} - 4 x^{3} - 18 x^{2} - 36 x + 441 = 0$

TS EAMCET-04.08.2021

Complex Numbers and Quadratic Equation

118476 If $\frac{α}{α + 1}$ and $\frac{β}{β + 1}$ are the roots of the quadratic equation $x^{2} + 7 x + 3 = 0$ , then the equation having roots $α$ and $β$ is

1

3 x^{2} - x - 3 = 0

2

11 x^{2} + 13 x + 3 = 0

3

13 x^{2} + 11 x + 13 = 0

4

11 x^{2} + 3 x + 13 = 0

Explanation:

B Let, $f (x) = x^{2} + 7 x + 3 = 0$
Roots are $\frac{α}{α + 1}$ and $\frac{β}{β + 1}$
Equation having roots $α$ and $β$
$Is, f (\frac{x}{x + 1}) = 0$
${(\frac{x}{x + 1})}^{2} + 7 (\frac{x}{x + 1}) + 3 = 0$
$x^{2} + 7 x (x + 1) + 3 (x + 1)^{2} = 0$
$x^{2} + 7 x^{2} + 7 x + 3 (x^{2} + 1 + 2 x) = 0$
$x^{2} + 7 x^{2} + 7 x + 3 x^{2} + 3 + 6 x = 0$
$11 x^{2} + 13 x + 3 = 0$

TS EAMCET-04.08.2021

Complex Numbers and Quadratic Equation

118473 The equation of lowest degree with rational coefficients having roots $\sqrt{3} + \sqrt{2} i$ and $\sqrt{3} - \sqrt{2}$ is

1

(x^{4} - 2 x^{2} + 25) (x^{4} - 10 x^{2} + 1) = 0

2

(x^{2} - 2 \sqrt{3} x + 5) (x^{2} - 2 \sqrt{3} x + 1) = 0

3

(x^{4} - 2 x^{2} + 25) (x^{4} + 10 x^{2} + 1) = 0

4

(x^{4} - 10 x^{2} + 1) (x^{4} + 2 x^{2} + 25) = 0

Explanation:

A The roots of the given equations are
$\sqrt{3} + \sqrt{2} i and \sqrt{3} - \sqrt{2}$
$x = \sqrt{3} + \sqrt{2} i$
$x - \sqrt{3} = \sqrt{2} i$
On squaring both sides
$x^{2} + 3 - 2 \sqrt{3} x = 2 i^{2}$
$x^{2} + 3 - 2 \sqrt{3} x = - 2$
$x^{2} + 5 = 2 \sqrt{3} x$
Again on squaring both sides
$x^{4} + 25 + 10 x^{2} = 12 x^{2}$
$x^{4} - 2 x^{2} + 25 = 0$
Also, $x = \sqrt{3} - \sqrt{2}$
On squaring both sides
$x^{2} = (\sqrt{3} - \sqrt{2})^{2}$
$x^{2} = 3 + 2 - 2 \sqrt{6}$
$x^{2} - 5 = - 2 \sqrt{6}$
Again on squaring both sides
${(x^{2} - 5)}^{2} = (- 2 \sqrt{6})^{2}$
$x^{4} + 25 - 10 x^{2} = 24$
$x^{4} - 10 x^{2} + 1 = 0$
$∴$ The required equation is
$(x^{4} - 2 x^{2} + 25) (x^{4} - 10 x^{2} + 1) = 0$

TS EAMCET-19.07.2022

Complex Numbers and Quadratic Equation

118474 $p$ is non-zero real number. If the equation whose roots are the squares of the roots of the equation $x^{3} - p x^{2} + p x - 1 = 0$ is identical with the given equation, then $p =$

1

\frac{1}{2}

2 2

3 3

4 -1

Explanation:

C Let $α, β, γ$ are roots of equation
$x^{3} - {px}^{2} + px - 1 = 0$
$α + β + γ = p$
$α β + β γ + α γ = p$
$α β γ = 1$
Also, given,
$α^{2} + β^{2} + γ^{2} = p$
$α^{2} β^{2} + β^{2} γ^{2} + α^{2} γ^{2} = p$
$α^{2} β^{2} γ^{2} = 1$
$∴ (α + β + γ)^{2} = α^{2} + β^{2} + γ^{2} + 2 (α β + β γ + γ α)$
$p^{2} = p + 2 p$
$p^{2} = 3 p$
$p = 3$

TS EAMCET-10.09.2020

Complex Numbers and Quadratic Equation

118475 The equation whose roots are squares of the roots of $x^{4} - 2 x^{3} + 6 x - 21 = 0$ is

1

x^{4} - 4 x^{3} - 18 x^{2} - 36 x + 441 = 0

2

x^{4} + 18 x^{3} - 4 x^{2} + 36 x + 441 = 0

3

x^{4} - 2 x^{3} + 4 x^{2} + 6 x + 441 = 0

4

x^{4} + 3 x^{3} - 5 x^{2} + 6 x + 441 = 0

Explanation:

Exp: (A): Let, $f (x) = x^{4} - 2 x^{3} + 6 x - 21 = 0$
Required equation where roots are squares of the roots of $f (x)$ is.
$f (\sqrt{x}) = 0$
$(\sqrt{x})^{4} - 2 (\sqrt{x})^{3} + 6 \sqrt{x} - 21 = 0$
$\Rightarrow x^{2} - 2 x \sqrt{x} + 6 \sqrt{x} - 21 = 0$
$\Rightarrow x^{2} - 21 = 2 x \sqrt{x} - 6 \sqrt{x}$
$\Rightarrow x^{2} - 21 = 2 \sqrt{x} (x - 3)$
$\Rightarrow x^{2} - 21 = 2 \sqrt{x} (x - 3)$
$\Rightarrow {(x^{2} - 21)}^{2} = 4 x (x - 3)^{2}$
$\Rightarrow x^{4} - 42 x^{2} + 441 = 4 x (x^{2} - 6 x + 9)$
$\Rightarrow x^{4} - 42 x^{2} + 441 = 4 x^{3} - 24 x^{2} + 36 x$
$\Rightarrow x^{4} - 4 x^{3} - 18 x^{2} - 36 x + 441 = 0$

TS EAMCET-04.08.2021

Complex Numbers and Quadratic Equation

118476 If $\frac{α}{α + 1}$ and $\frac{β}{β + 1}$ are the roots of the quadratic equation $x^{2} + 7 x + 3 = 0$ , then the equation having roots $α$ and $β$ is

1

3 x^{2} - x - 3 = 0

2

11 x^{2} + 13 x + 3 = 0

3

13 x^{2} + 11 x + 13 = 0

4

11 x^{2} + 3 x + 13 = 0

Explanation:

B Let, $f (x) = x^{2} + 7 x + 3 = 0$
Roots are $\frac{α}{α + 1}$ and $\frac{β}{β + 1}$
Equation having roots $α$ and $β$
$Is, f (\frac{x}{x + 1}) = 0$
${(\frac{x}{x + 1})}^{2} + 7 (\frac{x}{x + 1}) + 3 = 0$
$x^{2} + 7 x (x + 1) + 3 (x + 1)^{2} = 0$
$x^{2} + 7 x^{2} + 7 x + 3 (x^{2} + 1 + 2 x) = 0$
$x^{2} + 7 x^{2} + 7 x + 3 x^{2} + 3 + 6 x = 0$
$11 x^{2} + 13 x + 3 = 0$

TS EAMCET-04.08.2021

Complex Numbers and Quadratic Equation

118473 The equation of lowest degree with rational coefficients having roots $\sqrt{3} + \sqrt{2} i$ and $\sqrt{3} - \sqrt{2}$ is

1

(x^{4} - 2 x^{2} + 25) (x^{4} - 10 x^{2} + 1) = 0

2

(x^{2} - 2 \sqrt{3} x + 5) (x^{2} - 2 \sqrt{3} x + 1) = 0

3

(x^{4} - 2 x^{2} + 25) (x^{4} + 10 x^{2} + 1) = 0

4

(x^{4} - 10 x^{2} + 1) (x^{4} + 2 x^{2} + 25) = 0

Explanation:

A The roots of the given equations are
$\sqrt{3} + \sqrt{2} i and \sqrt{3} - \sqrt{2}$
$x = \sqrt{3} + \sqrt{2} i$
$x - \sqrt{3} = \sqrt{2} i$
On squaring both sides
$x^{2} + 3 - 2 \sqrt{3} x = 2 i^{2}$
$x^{2} + 3 - 2 \sqrt{3} x = - 2$
$x^{2} + 5 = 2 \sqrt{3} x$
Again on squaring both sides
$x^{4} + 25 + 10 x^{2} = 12 x^{2}$
$x^{4} - 2 x^{2} + 25 = 0$
Also, $x = \sqrt{3} - \sqrt{2}$
On squaring both sides
$x^{2} = (\sqrt{3} - \sqrt{2})^{2}$
$x^{2} = 3 + 2 - 2 \sqrt{6}$
$x^{2} - 5 = - 2 \sqrt{6}$
Again on squaring both sides
${(x^{2} - 5)}^{2} = (- 2 \sqrt{6})^{2}$
$x^{4} + 25 - 10 x^{2} = 24$
$x^{4} - 10 x^{2} + 1 = 0$
$∴$ The required equation is
$(x^{4} - 2 x^{2} + 25) (x^{4} - 10 x^{2} + 1) = 0$

TS EAMCET-19.07.2022

Complex Numbers and Quadratic Equation

118474 $p$ is non-zero real number. If the equation whose roots are the squares of the roots of the equation $x^{3} - p x^{2} + p x - 1 = 0$ is identical with the given equation, then $p =$

1

\frac{1}{2}

2 2

3 3

4 -1

Explanation:

C Let $α, β, γ$ are roots of equation
$x^{3} - {px}^{2} + px - 1 = 0$
$α + β + γ = p$
$α β + β γ + α γ = p$
$α β γ = 1$
Also, given,
$α^{2} + β^{2} + γ^{2} = p$
$α^{2} β^{2} + β^{2} γ^{2} + α^{2} γ^{2} = p$
$α^{2} β^{2} γ^{2} = 1$
$∴ (α + β + γ)^{2} = α^{2} + β^{2} + γ^{2} + 2 (α β + β γ + γ α)$
$p^{2} = p + 2 p$
$p^{2} = 3 p$
$p = 3$

TS EAMCET-10.09.2020

Complex Numbers and Quadratic Equation

118475 The equation whose roots are squares of the roots of $x^{4} - 2 x^{3} + 6 x - 21 = 0$ is

1

x^{4} - 4 x^{3} - 18 x^{2} - 36 x + 441 = 0

2

x^{4} + 18 x^{3} - 4 x^{2} + 36 x + 441 = 0

3

x^{4} - 2 x^{3} + 4 x^{2} + 6 x + 441 = 0

4

x^{4} + 3 x^{3} - 5 x^{2} + 6 x + 441 = 0

Explanation:

Exp: (A): Let, $f (x) = x^{4} - 2 x^{3} + 6 x - 21 = 0$
Required equation where roots are squares of the roots of $f (x)$ is.
$f (\sqrt{x}) = 0$
$(\sqrt{x})^{4} - 2 (\sqrt{x})^{3} + 6 \sqrt{x} - 21 = 0$
$\Rightarrow x^{2} - 2 x \sqrt{x} + 6 \sqrt{x} - 21 = 0$
$\Rightarrow x^{2} - 21 = 2 x \sqrt{x} - 6 \sqrt{x}$
$\Rightarrow x^{2} - 21 = 2 \sqrt{x} (x - 3)$
$\Rightarrow x^{2} - 21 = 2 \sqrt{x} (x - 3)$
$\Rightarrow {(x^{2} - 21)}^{2} = 4 x (x - 3)^{2}$
$\Rightarrow x^{4} - 42 x^{2} + 441 = 4 x (x^{2} - 6 x + 9)$
$\Rightarrow x^{4} - 42 x^{2} + 441 = 4 x^{3} - 24 x^{2} + 36 x$
$\Rightarrow x^{4} - 4 x^{3} - 18 x^{2} - 36 x + 441 = 0$

TS EAMCET-04.08.2021

Complex Numbers and Quadratic Equation

118476 If $\frac{α}{α + 1}$ and $\frac{β}{β + 1}$ are the roots of the quadratic equation $x^{2} + 7 x + 3 = 0$ , then the equation having roots $α$ and $β$ is

1

3 x^{2} - x - 3 = 0

2

11 x^{2} + 13 x + 3 = 0

3

13 x^{2} + 11 x + 13 = 0

4

11 x^{2} + 3 x + 13 = 0

Explanation:

B Let, $f (x) = x^{2} + 7 x + 3 = 0$
Roots are $\frac{α}{α + 1}$ and $\frac{β}{β + 1}$
Equation having roots $α$ and $β$
$Is, f (\frac{x}{x + 1}) = 0$
${(\frac{x}{x + 1})}^{2} + 7 (\frac{x}{x + 1}) + 3 = 0$
$x^{2} + 7 x (x + 1) + 3 (x + 1)^{2} = 0$
$x^{2} + 7 x^{2} + 7 x + 3 (x^{2} + 1 + 2 x) = 0$
$x^{2} + 7 x^{2} + 7 x + 3 x^{2} + 3 + 6 x = 0$
$11 x^{2} + 13 x + 3 = 0$

TS EAMCET-04.08.2021

Question Bank Series

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