QBManager

Complex Numbers and Quadratic Equation

118058 The real root of the equation $x^3-6 x+9=0$ is

1 6

2 -3

3 -6

4 -9

Karnataka CET-2008

Complex Numbers and Quadratic Equation

118060 If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3-8 x+8=0$, then $\quad \Sigma \alpha^2$ and $Σ \frac{1}{α β}$ are respectively

1 0 and -16

2 16 and 8

3 -16 and 0

4 16 and 0

Explanation:

D Given
$x^{3} - 8 x + 8 = 0$
$Then, α + β + γ = 0, α β + β γ + α γ = \frac{- 8}{1}$
$α β γ = \frac{- 8}{1}$
$\sum α^{2} = α^{2} + β^{2} + γ^{2} = (α + β + γ)^{2} - 2 (α β + β γ + α γ)$
$= 0^{2} - 2 \times \frac{- 8}{1} = 16$
$\sum \frac{1}{α β} = \frac{1}{α β} + \frac{1}{β γ} + \frac{1}{γ α} = \frac{α + β + γ}{α β γ} = \frac{0}{- 8} = 0$

Karnataka CET-2006

Complex Numbers and Quadratic Equation

118061 If $2 x = - 1 + \sqrt{3} i$ , then the value of
${(1 - x^{2} + x)}^{6} - {(1 - x + x^{2})}^{6} =$

1 32

2 -64

3 64

4 0

Explanation:

D Given, $2 x = - 1 + \sqrt{3} i$
Then, $x = - \frac{1}{2} + \frac{\sqrt{3} i}{2}$
We know that, $ω = - \frac{1}{2} + \frac{\sqrt{3} i}{2}$
Where $ω$ is cube root of unity
$∴ x = ω = \frac{- 1}{2} + \frac{\sqrt{3} i}{2}$
Now, ${(1 - x^{2} + x)}^{\frac{2}{6}} - {(1 - x + x^{2})}^{6}$
$= {(1 - ω^{2} + ω)}^{6} - {(1 - ω + ω^{2})}^{6}$
$= {(- ω^{2} - ω^{2})}^{6} - (- ω - ω)^{6} {∵ 1 + ω + ω^{2} = 0}$
$= {(- 2 ω^{2})}^{6} - (- 2 ω)^{6}$
$= 2^{6} - 2^{6} = 0 {∵ ω^{3} = 1}$

Karnataka CET-2006

Complex Numbers and Quadratic Equation

118062 If $α, β, γ$ are the roots of the equation $x^{3} - 3 x^{2} + 2 x - 1 = 0$ then the value of $[(1 - \boldsymbol α) (1 - \boldsymbol β) (1 - γ)]$ is

1 1

2 2

3 -1

4 -2

Explanation:

C Given, equation:
$x^{3} - 3 x^{2} + 2 x - 1 = 0$
$α + β + γ = 3,$
$α β + β γ + γ α = 2 and α β γ = 1$
$Now, (1 - α) (1 - β) (1 - γ)$
$= (1 - β - α + α β) (1 - γ)$
$= (1 - β - α + α β - γ + β γ + α γ - α β γ)$
$= [1 - (α + β + γ) + (α β + β γ + γ α) - α β γ]$
$= [1 - 3 + 2 - 1] = - 1$ Now, $(1 - α) (1 - β) (1 - γ)$

COMEDK-2012

Complex Numbers and Quadratic Equation

118058 The real root of the equation $x^{3} - 6 x + 9 = 0$ is

1 6

2 -3

3 -6

4 -9

Explanation:

B Given, $x^{3} - 6 x + 9 = 0$
Now, put $x = - 3$
$(- 3)^{3} - 6 (- 3) + 9$
$= - 27 + 18 + 9 = 0$
$∴ x = - 3$ is root of ${eq}^{n} x^{3} - 6 x + 9 = 0$

Karnataka CET-2008

Complex Numbers and Quadratic Equation

118060 If $α, β$ and $γ$ are the roots of the equation $x^{3} - 8 x + 8 = 0$ , then $Σ α^{2}$ and $Σ \frac{1}{α β}$ are respectively

1 0 and -16

2 16 and 8

3 -16 and 0

4 16 and 0

Explanation:

D Given
$x^{3} - 8 x + 8 = 0$
$Then, α + β + γ = 0, α β + β γ + α γ = \frac{- 8}{1}$
$α β γ = \frac{- 8}{1}$
$\sum α^{2} = α^{2} + β^{2} + γ^{2} = (α + β + γ)^{2} - 2 (α β + β γ + α γ)$
$= 0^{2} - 2 \times \frac{- 8}{1} = 16$
$\sum \frac{1}{α β} = \frac{1}{α β} + \frac{1}{β γ} + \frac{1}{γ α} = \frac{α + β + γ}{α β γ} = \frac{0}{- 8} = 0$

Karnataka CET-2006

Complex Numbers and Quadratic Equation

118061 If $2 x = - 1 + \sqrt{3} i$ , then the value of
${(1 - x^{2} + x)}^{6} - {(1 - x + x^{2})}^{6} =$

1 32

2 -64

3 64

4 0

Explanation:

D Given, $2 x = - 1 + \sqrt{3} i$
Then, $x = - \frac{1}{2} + \frac{\sqrt{3} i}{2}$
We know that, $ω = - \frac{1}{2} + \frac{\sqrt{3} i}{2}$
Where $ω$ is cube root of unity
$∴ x = ω = \frac{- 1}{2} + \frac{\sqrt{3} i}{2}$
Now, ${(1 - x^{2} + x)}^{\frac{2}{6}} - {(1 - x + x^{2})}^{6}$
$= {(1 - ω^{2} + ω)}^{6} - {(1 - ω + ω^{2})}^{6}$
$= {(- ω^{2} - ω^{2})}^{6} - (- ω - ω)^{6} {∵ 1 + ω + ω^{2} = 0}$
$= {(- 2 ω^{2})}^{6} - (- 2 ω)^{6}$
$= 2^{6} - 2^{6} = 0 {∵ ω^{3} = 1}$

Karnataka CET-2006

Complex Numbers and Quadratic Equation

118062 If $α, β, γ$ are the roots of the equation $x^{3} - 3 x^{2} + 2 x - 1 = 0$ then the value of $[(1 - \boldsymbol α) (1 - \boldsymbol β) (1 - γ)]$ is

1 1

2 2

3 -1

4 -2

Explanation:

C Given, equation:
$x^{3} - 3 x^{2} + 2 x - 1 = 0$
$α + β + γ = 3,$
$α β + β γ + γ α = 2 and α β γ = 1$
$Now, (1 - α) (1 - β) (1 - γ)$
$= (1 - β - α + α β) (1 - γ)$
$= (1 - β - α + α β - γ + β γ + α γ - α β γ)$
$= [1 - (α + β + γ) + (α β + β γ + γ α) - α β γ]$
$= [1 - 3 + 2 - 1] = - 1$ Now, $(1 - α) (1 - β) (1 - γ)$

COMEDK-2012

Complex Numbers and Quadratic Equation

118058 The real root of the equation $x^{3} - 6 x + 9 = 0$ is

1 6

2 -3

3 -6

4 -9

Explanation:

B Given, $x^{3} - 6 x + 9 = 0$
Now, put $x = - 3$
$(- 3)^{3} - 6 (- 3) + 9$
$= - 27 + 18 + 9 = 0$
$∴ x = - 3$ is root of ${eq}^{n} x^{3} - 6 x + 9 = 0$

Karnataka CET-2008

Complex Numbers and Quadratic Equation

118060 If $α, β$ and $γ$ are the roots of the equation $x^{3} - 8 x + 8 = 0$ , then $Σ α^{2}$ and $Σ \frac{1}{α β}$ are respectively

1 0 and -16

2 16 and 8

3 -16 and 0

4 16 and 0

Explanation:

D Given
$x^{3} - 8 x + 8 = 0$
$Then, α + β + γ = 0, α β + β γ + α γ = \frac{- 8}{1}$
$α β γ = \frac{- 8}{1}$
$\sum α^{2} = α^{2} + β^{2} + γ^{2} = (α + β + γ)^{2} - 2 (α β + β γ + α γ)$
$= 0^{2} - 2 \times \frac{- 8}{1} = 16$
$\sum \frac{1}{α β} = \frac{1}{α β} + \frac{1}{β γ} + \frac{1}{γ α} = \frac{α + β + γ}{α β γ} = \frac{0}{- 8} = 0$

Karnataka CET-2006

Complex Numbers and Quadratic Equation

118061 If $2 x = - 1 + \sqrt{3} i$ , then the value of
${(1 - x^{2} + x)}^{6} - {(1 - x + x^{2})}^{6} =$

1 32

2 -64

3 64

4 0

Explanation:

D Given, $2 x = - 1 + \sqrt{3} i$
Then, $x = - \frac{1}{2} + \frac{\sqrt{3} i}{2}$
We know that, $ω = - \frac{1}{2} + \frac{\sqrt{3} i}{2}$
Where $ω$ is cube root of unity
$∴ x = ω = \frac{- 1}{2} + \frac{\sqrt{3} i}{2}$
Now, ${(1 - x^{2} + x)}^{\frac{2}{6}} - {(1 - x + x^{2})}^{6}$
$= {(1 - ω^{2} + ω)}^{6} - {(1 - ω + ω^{2})}^{6}$
$= {(- ω^{2} - ω^{2})}^{6} - (- ω - ω)^{6} {∵ 1 + ω + ω^{2} = 0}$
$= {(- 2 ω^{2})}^{6} - (- 2 ω)^{6}$
$= 2^{6} - 2^{6} = 0 {∵ ω^{3} = 1}$

Karnataka CET-2006

Complex Numbers and Quadratic Equation

118062 If $α, β, γ$ are the roots of the equation $x^{3} - 3 x^{2} + 2 x - 1 = 0$ then the value of $[(1 - \boldsymbol α) (1 - \boldsymbol β) (1 - γ)]$ is

1 1

2 2

3 -1

4 -2

Explanation:

C Given, equation:
$x^{3} - 3 x^{2} + 2 x - 1 = 0$
$α + β + γ = 3,$
$α β + β γ + γ α = 2 and α β γ = 1$
$Now, (1 - α) (1 - β) (1 - γ)$
$= (1 - β - α + α β) (1 - γ)$
$= (1 - β - α + α β - γ + β γ + α γ - α β γ)$
$= [1 - (α + β + γ) + (α β + β γ + γ α) - α β γ]$
$= [1 - 3 + 2 - 1] = - 1$ Now, $(1 - α) (1 - β) (1 - γ)$

COMEDK-2012

Complex Numbers and Quadratic Equation

118058 The real root of the equation $x^{3} - 6 x + 9 = 0$ is

1 6

2 -3

3 -6

4 -9

Explanation:

B Given, $x^{3} - 6 x + 9 = 0$
Now, put $x = - 3$
$(- 3)^{3} - 6 (- 3) + 9$
$= - 27 + 18 + 9 = 0$
$∴ x = - 3$ is root of ${eq}^{n} x^{3} - 6 x + 9 = 0$

Karnataka CET-2008

Complex Numbers and Quadratic Equation

118060 If $α, β$ and $γ$ are the roots of the equation $x^{3} - 8 x + 8 = 0$ , then $Σ α^{2}$ and $Σ \frac{1}{α β}$ are respectively

1 0 and -16

2 16 and 8

3 -16 and 0

4 16 and 0

Explanation:

D Given
$x^{3} - 8 x + 8 = 0$
$Then, α + β + γ = 0, α β + β γ + α γ = \frac{- 8}{1}$
$α β γ = \frac{- 8}{1}$
$\sum α^{2} = α^{2} + β^{2} + γ^{2} = (α + β + γ)^{2} - 2 (α β + β γ + α γ)$
$= 0^{2} - 2 \times \frac{- 8}{1} = 16$
$\sum \frac{1}{α β} = \frac{1}{α β} + \frac{1}{β γ} + \frac{1}{γ α} = \frac{α + β + γ}{α β γ} = \frac{0}{- 8} = 0$

Karnataka CET-2006

Complex Numbers and Quadratic Equation

118061 If $2 x = - 1 + \sqrt{3} i$ , then the value of
${(1 - x^{2} + x)}^{6} - {(1 - x + x^{2})}^{6} =$

1 32

2 -64

3 64

4 0

Explanation:

D Given, $2 x = - 1 + \sqrt{3} i$
Then, $x = - \frac{1}{2} + \frac{\sqrt{3} i}{2}$
We know that, $ω = - \frac{1}{2} + \frac{\sqrt{3} i}{2}$
Where $ω$ is cube root of unity
$∴ x = ω = \frac{- 1}{2} + \frac{\sqrt{3} i}{2}$
Now, ${(1 - x^{2} + x)}^{\frac{2}{6}} - {(1 - x + x^{2})}^{6}$
$= {(1 - ω^{2} + ω)}^{6} - {(1 - ω + ω^{2})}^{6}$
$= {(- ω^{2} - ω^{2})}^{6} - (- ω - ω)^{6} {∵ 1 + ω + ω^{2} = 0}$
$= {(- 2 ω^{2})}^{6} - (- 2 ω)^{6}$
$= 2^{6} - 2^{6} = 0 {∵ ω^{3} = 1}$

Karnataka CET-2006

Complex Numbers and Quadratic Equation

118062 If $α, β, γ$ are the roots of the equation $x^{3} - 3 x^{2} + 2 x - 1 = 0$ then the value of $[(1 - \boldsymbol α) (1 - \boldsymbol β) (1 - γ)]$ is

1 1

2 2

3 -1

4 -2

Explanation:

C Given, equation:
$x^{3} - 3 x^{2} + 2 x - 1 = 0$
$α + β + γ = 3,$
$α β + β γ + γ α = 2 and α β γ = 1$
$Now, (1 - α) (1 - β) (1 - γ)$
$= (1 - β - α + α β) (1 - γ)$
$= (1 - β - α + α β - γ + β γ + α γ - α β γ)$
$= [1 - (α + β + γ) + (α β + β γ + γ α) - α β γ]$
$= [1 - 3 + 2 - 1] = - 1$ Now, $(1 - α) (1 - β) (1 - γ)$

COMEDK-2012

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: