QBManager

Complex Numbers and Quadratic Equation

118068 If $α, β$ are the roots of $x^{2} + p x + q = 0$ , and $ω$ is an imaginary cube root of unity, then value of $(ω α + ω^{2} β) (ω^{2} α + ω β)$ is

1

p^{2}

2

3 q

3

p^{2} - 2 q

4

p^{2} - 3 q

Explanation:

D $: Given equation: x^{2} + px + q = 0$
$Sum of roots (α + β) = - p$
$Product of roots (α β) = q$
$Now, (ω α + ω^{2} β) (ω^{2} α + ω β)$
$= α^{2} + β^{2} + (ω + ω^{2}) α β = α^{2} + β^{2} - α β {∵ ω + ω^{2} = - 1}$
$= (α + β)^{2} - 3 α β = p^{2} - 3 q$

BITSAT-2006

Complex Numbers and Quadratic Equation

118069 The number of real roots of
${(x + \frac{1}{x})}^{3} + (x + \frac{1}{x}) = 0$

1 0

2 2

3 4

4 6

Explanation:

A $W e h a v e, {(x + \frac{1}{x})}^{3} + (x + \frac{1}{x}) = 0$
$\Rightarrow (x + \frac{1}{x}) [{(x + \frac{1}{x})}^{2} + 1] = 0$
$\Rightarrow either x + \frac{1}{x} = 0 \Rightarrow x^{2} = - 1 \Rightarrow x = \pm i$
$or {(x + \frac{1}{x})}^{2} + 1 = 0$
$\Rightarrow x^{2} + \frac{1}{x^{2}} + 3 = 0 \Rightarrow x^{4} + 3 x^{2} + 1 = 0$
$\Rightarrow x^{2} = \frac{- 3 \pm \sqrt{9 - 4}}{2} = \frac{- 3 \pm \sqrt{5}}{2} < 0$
$∴ There is no real root.$ $∴$ There is no real root.

BITSAT-2005

Complex Numbers and Quadratic Equation

118071 If $x = ω - ω^{2} - 2$ , then the value of $x^{4} + 3 x^{3} + 2 x^{2} - 11 x - 6$ is

1 1

2 -1

3 2

4 None of these

Explanation:

A We have $x = ω - ω^{2} - 2$ or $x + 2 = ω - ω^{2}$
Squaring, $x^{2} + 4 x + 4 = ω^{2} + ω^{4} - 2 ω^{3}$
$= ω^{2} + ω^{3} ω - 2 ω^{3} = ω^{2} + ω - 2$
$= - 1 - 2 = - 3 \Rightarrow x^{2} + 4 x + 7 = 0$
Dividing $x^{4} + 3 x^{3} + 2 x^{2} - 11 x - 6$ by $x^{2} + 4 x + 7$
We get,
$x^{4} + 3 x^{3} + 2 x^{2} - 11 x - 6 = (x^{2} + 4 x + 7) (x^{2} - x - 1) + 1$
$= (0) (x^{2} - x - 1) + 1 = 0 + 1 = 1$

BITSAT-2012

Complex Numbers and Quadratic Equation

118073 The roots of the equation $x^{2} - 2 \sqrt{2} x + 1 = 0$ are

1 Real and different

2 Imaginary and different

3 Real and equal

4 Rational and different

Explanation:

A Given, equation $x^{2} - 2 \sqrt{2} x + 1 = 0$
The discriminant of the equation
$D > 0$
$(- 2 \sqrt{2})^{2} - 4 (1) (1) = 8 - 4 = 4 > 0$ and a perfect square
So, roots are real and different.

BITSAT-2010

Complex Numbers and Quadratic Equation

118068 If $α, β$ are the roots of $x^{2} + p x + q = 0$ , and $ω$ is an imaginary cube root of unity, then value of $(ω α + ω^{2} β) (ω^{2} α + ω β)$ is

1

p^{2}

2

3 q

3

p^{2} - 2 q

4

p^{2} - 3 q

Explanation:

D $: Given equation: x^{2} + px + q = 0$
$Sum of roots (α + β) = - p$
$Product of roots (α β) = q$
$Now, (ω α + ω^{2} β) (ω^{2} α + ω β)$
$= α^{2} + β^{2} + (ω + ω^{2}) α β = α^{2} + β^{2} - α β {∵ ω + ω^{2} = - 1}$
$= (α + β)^{2} - 3 α β = p^{2} - 3 q$

BITSAT-2006

Complex Numbers and Quadratic Equation

118069 The number of real roots of
${(x + \frac{1}{x})}^{3} + (x + \frac{1}{x}) = 0$

1 0

2 2

3 4

4 6

Explanation:

A $W e h a v e, {(x + \frac{1}{x})}^{3} + (x + \frac{1}{x}) = 0$
$\Rightarrow (x + \frac{1}{x}) [{(x + \frac{1}{x})}^{2} + 1] = 0$
$\Rightarrow either x + \frac{1}{x} = 0 \Rightarrow x^{2} = - 1 \Rightarrow x = \pm i$
$or {(x + \frac{1}{x})}^{2} + 1 = 0$
$\Rightarrow x^{2} + \frac{1}{x^{2}} + 3 = 0 \Rightarrow x^{4} + 3 x^{2} + 1 = 0$
$\Rightarrow x^{2} = \frac{- 3 \pm \sqrt{9 - 4}}{2} = \frac{- 3 \pm \sqrt{5}}{2} < 0$
$∴ There is no real root.$ $∴$ There is no real root.

BITSAT-2005

Complex Numbers and Quadratic Equation

118071 If $x = ω - ω^{2} - 2$ , then the value of $x^{4} + 3 x^{3} + 2 x^{2} - 11 x - 6$ is

1 1

2 -1

3 2

4 None of these

Explanation:

A We have $x = ω - ω^{2} - 2$ or $x + 2 = ω - ω^{2}$
Squaring, $x^{2} + 4 x + 4 = ω^{2} + ω^{4} - 2 ω^{3}$
$= ω^{2} + ω^{3} ω - 2 ω^{3} = ω^{2} + ω - 2$
$= - 1 - 2 = - 3 \Rightarrow x^{2} + 4 x + 7 = 0$
Dividing $x^{4} + 3 x^{3} + 2 x^{2} - 11 x - 6$ by $x^{2} + 4 x + 7$
We get,
$x^{4} + 3 x^{3} + 2 x^{2} - 11 x - 6 = (x^{2} + 4 x + 7) (x^{2} - x - 1) + 1$
$= (0) (x^{2} - x - 1) + 1 = 0 + 1 = 1$

BITSAT-2012

Complex Numbers and Quadratic Equation

118073 The roots of the equation $x^{2} - 2 \sqrt{2} x + 1 = 0$ are

1 Real and different

2 Imaginary and different

3 Real and equal

4 Rational and different

Explanation:

A Given, equation $x^{2} - 2 \sqrt{2} x + 1 = 0$
The discriminant of the equation
$D > 0$
$(- 2 \sqrt{2})^{2} - 4 (1) (1) = 8 - 4 = 4 > 0$ and a perfect square
So, roots are real and different.

BITSAT-2010

Complex Numbers and Quadratic Equation

118068 If $α, β$ are the roots of $x^{2} + p x + q = 0$ , and $ω$ is an imaginary cube root of unity, then value of $(ω α + ω^{2} β) (ω^{2} α + ω β)$ is

1

p^{2}

2

3 q

3

p^{2} - 2 q

4

p^{2} - 3 q

Explanation:

D $: Given equation: x^{2} + px + q = 0$
$Sum of roots (α + β) = - p$
$Product of roots (α β) = q$
$Now, (ω α + ω^{2} β) (ω^{2} α + ω β)$
$= α^{2} + β^{2} + (ω + ω^{2}) α β = α^{2} + β^{2} - α β {∵ ω + ω^{2} = - 1}$
$= (α + β)^{2} - 3 α β = p^{2} - 3 q$

BITSAT-2006

Complex Numbers and Quadratic Equation

118069 The number of real roots of
${(x + \frac{1}{x})}^{3} + (x + \frac{1}{x}) = 0$

1 0

2 2

3 4

4 6

Explanation:

A $W e h a v e, {(x + \frac{1}{x})}^{3} + (x + \frac{1}{x}) = 0$
$\Rightarrow (x + \frac{1}{x}) [{(x + \frac{1}{x})}^{2} + 1] = 0$
$\Rightarrow either x + \frac{1}{x} = 0 \Rightarrow x^{2} = - 1 \Rightarrow x = \pm i$
$or {(x + \frac{1}{x})}^{2} + 1 = 0$
$\Rightarrow x^{2} + \frac{1}{x^{2}} + 3 = 0 \Rightarrow x^{4} + 3 x^{2} + 1 = 0$
$\Rightarrow x^{2} = \frac{- 3 \pm \sqrt{9 - 4}}{2} = \frac{- 3 \pm \sqrt{5}}{2} < 0$
$∴ There is no real root.$ $∴$ There is no real root.

BITSAT-2005

Complex Numbers and Quadratic Equation

118071 If $x = ω - ω^{2} - 2$ , then the value of $x^{4} + 3 x^{3} + 2 x^{2} - 11 x - 6$ is

1 1

2 -1

3 2

4 None of these

Explanation:

A We have $x = ω - ω^{2} - 2$ or $x + 2 = ω - ω^{2}$
Squaring, $x^{2} + 4 x + 4 = ω^{2} + ω^{4} - 2 ω^{3}$
$= ω^{2} + ω^{3} ω - 2 ω^{3} = ω^{2} + ω - 2$
$= - 1 - 2 = - 3 \Rightarrow x^{2} + 4 x + 7 = 0$
Dividing $x^{4} + 3 x^{3} + 2 x^{2} - 11 x - 6$ by $x^{2} + 4 x + 7$
We get,
$x^{4} + 3 x^{3} + 2 x^{2} - 11 x - 6 = (x^{2} + 4 x + 7) (x^{2} - x - 1) + 1$
$= (0) (x^{2} - x - 1) + 1 = 0 + 1 = 1$

BITSAT-2012

Complex Numbers and Quadratic Equation

118073 The roots of the equation $x^{2} - 2 \sqrt{2} x + 1 = 0$ are

1 Real and different

2 Imaginary and different

3 Real and equal

4 Rational and different

Explanation:

A Given, equation $x^{2} - 2 \sqrt{2} x + 1 = 0$
The discriminant of the equation
$D > 0$
$(- 2 \sqrt{2})^{2} - 4 (1) (1) = 8 - 4 = 4 > 0$ and a perfect square
So, roots are real and different.

BITSAT-2010

Complex Numbers and Quadratic Equation

118068 If $α, β$ are the roots of $x^{2} + p x + q = 0$ , and $ω$ is an imaginary cube root of unity, then value of $(ω α + ω^{2} β) (ω^{2} α + ω β)$ is

1

p^{2}

2

3 q

3

p^{2} - 2 q

4

p^{2} - 3 q

Explanation:

D $: Given equation: x^{2} + px + q = 0$
$Sum of roots (α + β) = - p$
$Product of roots (α β) = q$
$Now, (ω α + ω^{2} β) (ω^{2} α + ω β)$
$= α^{2} + β^{2} + (ω + ω^{2}) α β = α^{2} + β^{2} - α β {∵ ω + ω^{2} = - 1}$
$= (α + β)^{2} - 3 α β = p^{2} - 3 q$

BITSAT-2006

Complex Numbers and Quadratic Equation

118069 The number of real roots of
${(x + \frac{1}{x})}^{3} + (x + \frac{1}{x}) = 0$

1 0

2 2

3 4

4 6

Explanation:

A $W e h a v e, {(x + \frac{1}{x})}^{3} + (x + \frac{1}{x}) = 0$
$\Rightarrow (x + \frac{1}{x}) [{(x + \frac{1}{x})}^{2} + 1] = 0$
$\Rightarrow either x + \frac{1}{x} = 0 \Rightarrow x^{2} = - 1 \Rightarrow x = \pm i$
$or {(x + \frac{1}{x})}^{2} + 1 = 0$
$\Rightarrow x^{2} + \frac{1}{x^{2}} + 3 = 0 \Rightarrow x^{4} + 3 x^{2} + 1 = 0$
$\Rightarrow x^{2} = \frac{- 3 \pm \sqrt{9 - 4}}{2} = \frac{- 3 \pm \sqrt{5}}{2} < 0$
$∴ There is no real root.$ $∴$ There is no real root.

BITSAT-2005

Complex Numbers and Quadratic Equation

118071 If $x = ω - ω^{2} - 2$ , then the value of $x^{4} + 3 x^{3} + 2 x^{2} - 11 x - 6$ is

1 1

2 -1

3 2

4 None of these

Explanation:

A We have $x = ω - ω^{2} - 2$ or $x + 2 = ω - ω^{2}$
Squaring, $x^{2} + 4 x + 4 = ω^{2} + ω^{4} - 2 ω^{3}$
$= ω^{2} + ω^{3} ω - 2 ω^{3} = ω^{2} + ω - 2$
$= - 1 - 2 = - 3 \Rightarrow x^{2} + 4 x + 7 = 0$
Dividing $x^{4} + 3 x^{3} + 2 x^{2} - 11 x - 6$ by $x^{2} + 4 x + 7$
We get,
$x^{4} + 3 x^{3} + 2 x^{2} - 11 x - 6 = (x^{2} + 4 x + 7) (x^{2} - x - 1) + 1$
$= (0) (x^{2} - x - 1) + 1 = 0 + 1 = 1$

BITSAT-2012

Complex Numbers and Quadratic Equation

118073 The roots of the equation $x^{2} - 2 \sqrt{2} x + 1 = 0$ are

1 Real and different

2 Imaginary and different

3 Real and equal

4 Rational and different

Explanation:

A Given, equation $x^{2} - 2 \sqrt{2} x + 1 = 0$
The discriminant of the equation
$D > 0$
$(- 2 \sqrt{2})^{2} - 4 (1) (1) = 8 - 4 = 4 > 0$ and a perfect square
So, roots are real and different.

BITSAT-2010

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: