QBManager

Complex Numbers and Quadratic Equation

118130 If $α$ and $β$ are the roots of $x^{2} - 2 x + 4 = 0$ , then the value of $α^{6} + β^{6}$ is

1 32

2 64

3 128

4 256

Explanation:

C $α^{6} + β^{6} = {(α^{2})}^{3} + {(β^{2})}^{3}$
$= (α^{2} + β^{2}) ({(α^{2} + β^{2})}^{2} - 3 (α β)^{2})$
$We have the equation$
$x^{2} - 2 x + 4 = 0 and α, β are the roots of the eqn.$
$∴ α + β = 2$
$α β = 4$
$(α + β)^{2} = α^{2} + β^{2} + 2 α β$
$∴ α^{2} + β^{2} = (α + β)^{2} - 2 α β = 4 - 8 = - 4$
$∴ α^{6} + β^{6} = - 4 ((- 4)^{2} - 3 (4^{2})]$
$= - 4 (16 - 48)$
$= - 4 \times (- 32)$
$= 128$
$∴ Option (c) is correct.$

AP EAMCET-2009

Complex Numbers and Quadratic Equation

118131 If $α$ and $β$ are the roots of the equation $a x^{2} + b x$ $+ c = 0$ and, if ${p x}^{2} + q x + r = 0$ has roots $\frac{1 - α}{α}$ and $\frac{1 - β}{β}$ , then $r$ is equal to

1

a + 2 b

2

a + b + c

3

ab + bc + ca

4 abc

Explanation:

B $α and β are the roots of$
$a x^{2} + b x + c = 0$
$α + β = - b / a$
$α β = c / a$
$Given, {px}^{2} + qx + r = 0 has roots$
$\frac{1 - α}{α} and \frac{1 - β}{β}$
$\Rightarrow \frac{1 - α}{α} + \frac{1 - β}{β} = \frac{- q}{p}$
$and (\frac{1 - α}{α}) (\frac{1 - β}{β}) = \frac{r}{p}$
$or \frac{1}{α} + \frac{1}{β} = 2 - \frac{q}{p}$
$∴ \frac{(α + β)}{α β} = 2 - \frac{q}{p} = \frac{- b / a}{c / a} = \frac{- b}{c}$
$\frac{(1 - α) (1 - β)}{α β} = \frac{1 - α - β + α β}{α β}$
$= \frac{1 - (α + β) + α β}{α β}$
$= \frac{1 + \frac{b}{a} + \frac{c}{a}}{\frac{c}{a}} = \frac{a + b + c}{\frac{a}{c}} = \frac{a + b + c}{c}$
$\frac{r}{p} = \frac{a + b + c}{c}$
$r = a + b + c$
From eq $^{n}$ (i) $\frac{1}{α} - 1 + \frac{1}{β} - 1 = \frac{- q}{p}$
On comparing both side

AP EAMCET-2007

Complex Numbers and Quadratic Equation

118132 The set of values of $x$ for which the inequalities $x^{2} - 3 x - 10 < 0, 10 x - x^{2} - 16 > 0$ hold simultaneously, is

1

(- 2, 5)

2

(2, 8)

3

(- 2, 8)

4

(2, 5)

Explanation:

D Given $x^{2} - 3 x - 10 < 0 and$
$10 x - x^{2} - 16 > 0 hold.$
$∴ x^{2} - 3 x - 10 = x^{2} - 5 x + 2 x - 10$
$= x (x - 5) + 2 (x - 5)$
$= (x + 2) (x - 5) < 0$
$\Rightarrow - 2 < x < 5$
$Similarly, 10 x - x^{2} - 16$
$= - x^{2} + 10 x - 16 > 0$
$\Rightarrow x^{2} - 10 x + 16 < 0$
$\Rightarrow x^{2} - 8 x - 2 x + 16 < 0$
$\Rightarrow x (x - 8) - 2 (x - 8) < 0$
$\Rightarrow (x - 8) (x - 2) < 0$
$\Rightarrow 2 < x < 8$
$Both hold simultaneously ∴ 2 < x < 5$
$∴ x \in (2, 5)$
$∴ Option (d) is correct.$ Both hold simultaneously $∴ 2 < x < 5$

AP EAMCET-2007

Complex Numbers and Quadratic Equation

118164 If one root of the equation $x^{2} + p x + 12 = 0$ is 4 while the equation $x^{2} + p x + q = 0$ has equal roots, then the value of $q$ is

1

\frac{49}{4}

2 12

3 3

4 4

Explanation:

A Given one root of the equation
$x^{2} + px + 12 = 0$ is 4
$∴ (4)^{2} + 4 p + 12 = 0$
$\Rightarrow 4 p = - 12 - 16 = - 28$
$∴ p = \frac{- 28}{4} = - 7$
Also, given the equation $x^{2} + px + q = 0$ has equal roots $\Rightarrow D = 0, (p$ is criminate $) = 0$
i.e. $p^{2} - 4 q = 0$
$or (- 7)^{2} = 4 q$
$∴ q = \frac{49}{4}$
$∴$ Option (a) is correct.

AIEEE-2004

Complex Numbers and Quadratic Equation

118133 If $1, 2, 3$ and 4 are the roots of the equation $x^{4} +$ $a x + {b x}^{2} + c x + d = 0$ , then $a + 2 b + c$ is equal to

1 -25

2 0

3 10

4 24

Explanation:

C Given the equation
$x^{4} + a x^{3} + b x^{2} + cx + d = 0$
If $1, 2, 3$ and 4 are the roots of above
Then, $1 + a + b + c + d = 0$
$16 + 8 a + 4 b + 2 c + d = 0$
$81 + 27 a + a b + 3 c + d = 0$
$256 + 64 a 16 b + 4 c + d = 0$
Subtracting (i) from (ii) we get
$15 + 7 a + 3 b + c = 0$
$Subtracting (ii) from (iii)$
$65 + 19 a + 5 b + c = 0$
Subtracting (iii) from (iv)
$175 + 37 a + 7 b + c = 0$
Subtracting (v) from (vi), we get
$50 + 12 a + 2 b = 0$
Subtracting (vi) from (vii), we get
$110 + 18 a + 2 b = 0$
Subtracting (viii) and (ix), we get
$60 + 6 a = 0 \Rightarrow a = - 10$
$2 b = - 110 - 18 (- 10)$
$or 2 b = 70$
$∴ b = 35$
$∴ From eqn (v)$
$15 + 7 x (- 10) + 3 \times 35 + c = 0$
$or 15 - 70 + 105 + c = 0$
$or c = - 105 - 15 + 70$
$= - 120 + 70$
$= - 50$
$∴ a + 2 b + c = - 10 + 70 - 50$
$= 10$ $∴$ Option (c) is correct.

AP EAMCET-2007

Complex Numbers and Quadratic Equation

118130 If $α$ and $β$ are the roots of $x^{2} - 2 x + 4 = 0$ , then the value of $α^{6} + β^{6}$ is

1 32

2 64

3 128

4 256

Explanation:

C $α^{6} + β^{6} = {(α^{2})}^{3} + {(β^{2})}^{3}$
$= (α^{2} + β^{2}) ({(α^{2} + β^{2})}^{2} - 3 (α β)^{2})$
$We have the equation$
$x^{2} - 2 x + 4 = 0 and α, β are the roots of the eqn.$
$∴ α + β = 2$
$α β = 4$
$(α + β)^{2} = α^{2} + β^{2} + 2 α β$
$∴ α^{2} + β^{2} = (α + β)^{2} - 2 α β = 4 - 8 = - 4$
$∴ α^{6} + β^{6} = - 4 ((- 4)^{2} - 3 (4^{2})]$
$= - 4 (16 - 48)$
$= - 4 \times (- 32)$
$= 128$
$∴ Option (c) is correct.$

AP EAMCET-2009

Complex Numbers and Quadratic Equation

118131 If $α$ and $β$ are the roots of the equation $a x^{2} + b x$ $+ c = 0$ and, if ${p x}^{2} + q x + r = 0$ has roots $\frac{1 - α}{α}$ and $\frac{1 - β}{β}$ , then $r$ is equal to

1

a + 2 b

2

a + b + c

3

ab + bc + ca

4 abc

Explanation:

B $α and β are the roots of$
$a x^{2} + b x + c = 0$
$α + β = - b / a$
$α β = c / a$
$Given, {px}^{2} + qx + r = 0 has roots$
$\frac{1 - α}{α} and \frac{1 - β}{β}$
$\Rightarrow \frac{1 - α}{α} + \frac{1 - β}{β} = \frac{- q}{p}$
$and (\frac{1 - α}{α}) (\frac{1 - β}{β}) = \frac{r}{p}$
$or \frac{1}{α} + \frac{1}{β} = 2 - \frac{q}{p}$
$∴ \frac{(α + β)}{α β} = 2 - \frac{q}{p} = \frac{- b / a}{c / a} = \frac{- b}{c}$
$\frac{(1 - α) (1 - β)}{α β} = \frac{1 - α - β + α β}{α β}$
$= \frac{1 - (α + β) + α β}{α β}$
$= \frac{1 + \frac{b}{a} + \frac{c}{a}}{\frac{c}{a}} = \frac{a + b + c}{\frac{a}{c}} = \frac{a + b + c}{c}$
$\frac{r}{p} = \frac{a + b + c}{c}$
$r = a + b + c$
From eq $^{n}$ (i) $\frac{1}{α} - 1 + \frac{1}{β} - 1 = \frac{- q}{p}$
On comparing both side

AP EAMCET-2007

Complex Numbers and Quadratic Equation

118132 The set of values of $x$ for which the inequalities $x^{2} - 3 x - 10 < 0, 10 x - x^{2} - 16 > 0$ hold simultaneously, is

1

(- 2, 5)

2

(2, 8)

3

(- 2, 8)

4

(2, 5)

Explanation:

D Given $x^{2} - 3 x - 10 < 0 and$
$10 x - x^{2} - 16 > 0 hold.$
$∴ x^{2} - 3 x - 10 = x^{2} - 5 x + 2 x - 10$
$= x (x - 5) + 2 (x - 5)$
$= (x + 2) (x - 5) < 0$
$\Rightarrow - 2 < x < 5$
$Similarly, 10 x - x^{2} - 16$
$= - x^{2} + 10 x - 16 > 0$
$\Rightarrow x^{2} - 10 x + 16 < 0$
$\Rightarrow x^{2} - 8 x - 2 x + 16 < 0$
$\Rightarrow x (x - 8) - 2 (x - 8) < 0$
$\Rightarrow (x - 8) (x - 2) < 0$
$\Rightarrow 2 < x < 8$
$Both hold simultaneously ∴ 2 < x < 5$
$∴ x \in (2, 5)$
$∴ Option (d) is correct.$ Both hold simultaneously $∴ 2 < x < 5$

AP EAMCET-2007

Complex Numbers and Quadratic Equation

118164 If one root of the equation $x^{2} + p x + 12 = 0$ is 4 while the equation $x^{2} + p x + q = 0$ has equal roots, then the value of $q$ is

1

\frac{49}{4}

2 12

3 3

4 4

Explanation:

A Given one root of the equation
$x^{2} + px + 12 = 0$ is 4
$∴ (4)^{2} + 4 p + 12 = 0$
$\Rightarrow 4 p = - 12 - 16 = - 28$
$∴ p = \frac{- 28}{4} = - 7$
Also, given the equation $x^{2} + px + q = 0$ has equal roots $\Rightarrow D = 0, (p$ is criminate $) = 0$
i.e. $p^{2} - 4 q = 0$
$or (- 7)^{2} = 4 q$
$∴ q = \frac{49}{4}$
$∴$ Option (a) is correct.

AIEEE-2004

Complex Numbers and Quadratic Equation

118133 If $1, 2, 3$ and 4 are the roots of the equation $x^{4} +$ $a x + {b x}^{2} + c x + d = 0$ , then $a + 2 b + c$ is equal to

1 -25

2 0

3 10

4 24

Explanation:

C Given the equation
$x^{4} + a x^{3} + b x^{2} + cx + d = 0$
If $1, 2, 3$ and 4 are the roots of above
Then, $1 + a + b + c + d = 0$
$16 + 8 a + 4 b + 2 c + d = 0$
$81 + 27 a + a b + 3 c + d = 0$
$256 + 64 a 16 b + 4 c + d = 0$
Subtracting (i) from (ii) we get
$15 + 7 a + 3 b + c = 0$
$Subtracting (ii) from (iii)$
$65 + 19 a + 5 b + c = 0$
Subtracting (iii) from (iv)
$175 + 37 a + 7 b + c = 0$
Subtracting (v) from (vi), we get
$50 + 12 a + 2 b = 0$
Subtracting (vi) from (vii), we get
$110 + 18 a + 2 b = 0$
Subtracting (viii) and (ix), we get
$60 + 6 a = 0 \Rightarrow a = - 10$
$2 b = - 110 - 18 (- 10)$
$or 2 b = 70$
$∴ b = 35$
$∴ From eqn (v)$
$15 + 7 x (- 10) + 3 \times 35 + c = 0$
$or 15 - 70 + 105 + c = 0$
$or c = - 105 - 15 + 70$
$= - 120 + 70$
$= - 50$
$∴ a + 2 b + c = - 10 + 70 - 50$
$= 10$ $∴$ Option (c) is correct.

AP EAMCET-2007

Complex Numbers and Quadratic Equation

118130 If $α$ and $β$ are the roots of $x^{2} - 2 x + 4 = 0$ , then the value of $α^{6} + β^{6}$ is

1 32

2 64

3 128

4 256

Explanation:

C $α^{6} + β^{6} = {(α^{2})}^{3} + {(β^{2})}^{3}$
$= (α^{2} + β^{2}) ({(α^{2} + β^{2})}^{2} - 3 (α β)^{2})$
$We have the equation$
$x^{2} - 2 x + 4 = 0 and α, β are the roots of the eqn.$
$∴ α + β = 2$
$α β = 4$
$(α + β)^{2} = α^{2} + β^{2} + 2 α β$
$∴ α^{2} + β^{2} = (α + β)^{2} - 2 α β = 4 - 8 = - 4$
$∴ α^{6} + β^{6} = - 4 ((- 4)^{2} - 3 (4^{2})]$
$= - 4 (16 - 48)$
$= - 4 \times (- 32)$
$= 128$
$∴ Option (c) is correct.$

AP EAMCET-2009

Complex Numbers and Quadratic Equation

118131 If $α$ and $β$ are the roots of the equation $a x^{2} + b x$ $+ c = 0$ and, if ${p x}^{2} + q x + r = 0$ has roots $\frac{1 - α}{α}$ and $\frac{1 - β}{β}$ , then $r$ is equal to

1

a + 2 b

2

a + b + c

3

ab + bc + ca

4 abc

Explanation:

B $α and β are the roots of$
$a x^{2} + b x + c = 0$
$α + β = - b / a$
$α β = c / a$
$Given, {px}^{2} + qx + r = 0 has roots$
$\frac{1 - α}{α} and \frac{1 - β}{β}$
$\Rightarrow \frac{1 - α}{α} + \frac{1 - β}{β} = \frac{- q}{p}$
$and (\frac{1 - α}{α}) (\frac{1 - β}{β}) = \frac{r}{p}$
$or \frac{1}{α} + \frac{1}{β} = 2 - \frac{q}{p}$
$∴ \frac{(α + β)}{α β} = 2 - \frac{q}{p} = \frac{- b / a}{c / a} = \frac{- b}{c}$
$\frac{(1 - α) (1 - β)}{α β} = \frac{1 - α - β + α β}{α β}$
$= \frac{1 - (α + β) + α β}{α β}$
$= \frac{1 + \frac{b}{a} + \frac{c}{a}}{\frac{c}{a}} = \frac{a + b + c}{\frac{a}{c}} = \frac{a + b + c}{c}$
$\frac{r}{p} = \frac{a + b + c}{c}$
$r = a + b + c$
From eq $^{n}$ (i) $\frac{1}{α} - 1 + \frac{1}{β} - 1 = \frac{- q}{p}$
On comparing both side

AP EAMCET-2007

Complex Numbers and Quadratic Equation

118132 The set of values of $x$ for which the inequalities $x^{2} - 3 x - 10 < 0, 10 x - x^{2} - 16 > 0$ hold simultaneously, is

1

(- 2, 5)

2

(2, 8)

3

(- 2, 8)

4

(2, 5)

Explanation:

D Given $x^{2} - 3 x - 10 < 0 and$
$10 x - x^{2} - 16 > 0 hold.$
$∴ x^{2} - 3 x - 10 = x^{2} - 5 x + 2 x - 10$
$= x (x - 5) + 2 (x - 5)$
$= (x + 2) (x - 5) < 0$
$\Rightarrow - 2 < x < 5$
$Similarly, 10 x - x^{2} - 16$
$= - x^{2} + 10 x - 16 > 0$
$\Rightarrow x^{2} - 10 x + 16 < 0$
$\Rightarrow x^{2} - 8 x - 2 x + 16 < 0$
$\Rightarrow x (x - 8) - 2 (x - 8) < 0$
$\Rightarrow (x - 8) (x - 2) < 0$
$\Rightarrow 2 < x < 8$
$Both hold simultaneously ∴ 2 < x < 5$
$∴ x \in (2, 5)$
$∴ Option (d) is correct.$ Both hold simultaneously $∴ 2 < x < 5$

AP EAMCET-2007

Complex Numbers and Quadratic Equation

118164 If one root of the equation $x^{2} + p x + 12 = 0$ is 4 while the equation $x^{2} + p x + q = 0$ has equal roots, then the value of $q$ is

1

\frac{49}{4}

2 12

3 3

4 4

Explanation:

A Given one root of the equation
$x^{2} + px + 12 = 0$ is 4
$∴ (4)^{2} + 4 p + 12 = 0$
$\Rightarrow 4 p = - 12 - 16 = - 28$
$∴ p = \frac{- 28}{4} = - 7$
Also, given the equation $x^{2} + px + q = 0$ has equal roots $\Rightarrow D = 0, (p$ is criminate $) = 0$
i.e. $p^{2} - 4 q = 0$
$or (- 7)^{2} = 4 q$
$∴ q = \frac{49}{4}$
$∴$ Option (a) is correct.

AIEEE-2004

Complex Numbers and Quadratic Equation

118133 If $1, 2, 3$ and 4 are the roots of the equation $x^{4} +$ $a x + {b x}^{2} + c x + d = 0$ , then $a + 2 b + c$ is equal to

1 -25

2 0

3 10

4 24

Explanation:

C Given the equation
$x^{4} + a x^{3} + b x^{2} + cx + d = 0$
If $1, 2, 3$ and 4 are the roots of above
Then, $1 + a + b + c + d = 0$
$16 + 8 a + 4 b + 2 c + d = 0$
$81 + 27 a + a b + 3 c + d = 0$
$256 + 64 a 16 b + 4 c + d = 0$
Subtracting (i) from (ii) we get
$15 + 7 a + 3 b + c = 0$
$Subtracting (ii) from (iii)$
$65 + 19 a + 5 b + c = 0$
Subtracting (iii) from (iv)
$175 + 37 a + 7 b + c = 0$
Subtracting (v) from (vi), we get
$50 + 12 a + 2 b = 0$
Subtracting (vi) from (vii), we get
$110 + 18 a + 2 b = 0$
Subtracting (viii) and (ix), we get
$60 + 6 a = 0 \Rightarrow a = - 10$
$2 b = - 110 - 18 (- 10)$
$or 2 b = 70$
$∴ b = 35$
$∴ From eqn (v)$
$15 + 7 x (- 10) + 3 \times 35 + c = 0$
$or 15 - 70 + 105 + c = 0$
$or c = - 105 - 15 + 70$
$= - 120 + 70$
$= - 50$
$∴ a + 2 b + c = - 10 + 70 - 50$
$= 10$ $∴$ Option (c) is correct.

AP EAMCET-2007

Complex Numbers and Quadratic Equation

118130 If $α$ and $β$ are the roots of $x^{2} - 2 x + 4 = 0$ , then the value of $α^{6} + β^{6}$ is

1 32

2 64

3 128

4 256

Explanation:

C $α^{6} + β^{6} = {(α^{2})}^{3} + {(β^{2})}^{3}$
$= (α^{2} + β^{2}) ({(α^{2} + β^{2})}^{2} - 3 (α β)^{2})$
$We have the equation$
$x^{2} - 2 x + 4 = 0 and α, β are the roots of the eqn.$
$∴ α + β = 2$
$α β = 4$
$(α + β)^{2} = α^{2} + β^{2} + 2 α β$
$∴ α^{2} + β^{2} = (α + β)^{2} - 2 α β = 4 - 8 = - 4$
$∴ α^{6} + β^{6} = - 4 ((- 4)^{2} - 3 (4^{2})]$
$= - 4 (16 - 48)$
$= - 4 \times (- 32)$
$= 128$
$∴ Option (c) is correct.$

AP EAMCET-2009

Complex Numbers and Quadratic Equation

118131 If $α$ and $β$ are the roots of the equation $a x^{2} + b x$ $+ c = 0$ and, if ${p x}^{2} + q x + r = 0$ has roots $\frac{1 - α}{α}$ and $\frac{1 - β}{β}$ , then $r$ is equal to

1

a + 2 b

2

a + b + c

3

ab + bc + ca

4 abc

Explanation:

B $α and β are the roots of$
$a x^{2} + b x + c = 0$
$α + β = - b / a$
$α β = c / a$
$Given, {px}^{2} + qx + r = 0 has roots$
$\frac{1 - α}{α} and \frac{1 - β}{β}$
$\Rightarrow \frac{1 - α}{α} + \frac{1 - β}{β} = \frac{- q}{p}$
$and (\frac{1 - α}{α}) (\frac{1 - β}{β}) = \frac{r}{p}$
$or \frac{1}{α} + \frac{1}{β} = 2 - \frac{q}{p}$
$∴ \frac{(α + β)}{α β} = 2 - \frac{q}{p} = \frac{- b / a}{c / a} = \frac{- b}{c}$
$\frac{(1 - α) (1 - β)}{α β} = \frac{1 - α - β + α β}{α β}$
$= \frac{1 - (α + β) + α β}{α β}$
$= \frac{1 + \frac{b}{a} + \frac{c}{a}}{\frac{c}{a}} = \frac{a + b + c}{\frac{a}{c}} = \frac{a + b + c}{c}$
$\frac{r}{p} = \frac{a + b + c}{c}$
$r = a + b + c$
From eq $^{n}$ (i) $\frac{1}{α} - 1 + \frac{1}{β} - 1 = \frac{- q}{p}$
On comparing both side

AP EAMCET-2007

Complex Numbers and Quadratic Equation

118132 The set of values of $x$ for which the inequalities $x^{2} - 3 x - 10 < 0, 10 x - x^{2} - 16 > 0$ hold simultaneously, is

1

(- 2, 5)

2

(2, 8)

3

(- 2, 8)

4

(2, 5)

Explanation:

D Given $x^{2} - 3 x - 10 < 0 and$
$10 x - x^{2} - 16 > 0 hold.$
$∴ x^{2} - 3 x - 10 = x^{2} - 5 x + 2 x - 10$
$= x (x - 5) + 2 (x - 5)$
$= (x + 2) (x - 5) < 0$
$\Rightarrow - 2 < x < 5$
$Similarly, 10 x - x^{2} - 16$
$= - x^{2} + 10 x - 16 > 0$
$\Rightarrow x^{2} - 10 x + 16 < 0$
$\Rightarrow x^{2} - 8 x - 2 x + 16 < 0$
$\Rightarrow x (x - 8) - 2 (x - 8) < 0$
$\Rightarrow (x - 8) (x - 2) < 0$
$\Rightarrow 2 < x < 8$
$Both hold simultaneously ∴ 2 < x < 5$
$∴ x \in (2, 5)$
$∴ Option (d) is correct.$ Both hold simultaneously $∴ 2 < x < 5$

AP EAMCET-2007

Complex Numbers and Quadratic Equation

118164 If one root of the equation $x^{2} + p x + 12 = 0$ is 4 while the equation $x^{2} + p x + q = 0$ has equal roots, then the value of $q$ is

1

\frac{49}{4}

2 12

3 3

4 4

Explanation:

A Given one root of the equation
$x^{2} + px + 12 = 0$ is 4
$∴ (4)^{2} + 4 p + 12 = 0$
$\Rightarrow 4 p = - 12 - 16 = - 28$
$∴ p = \frac{- 28}{4} = - 7$
Also, given the equation $x^{2} + px + q = 0$ has equal roots $\Rightarrow D = 0, (p$ is criminate $) = 0$
i.e. $p^{2} - 4 q = 0$
$or (- 7)^{2} = 4 q$
$∴ q = \frac{49}{4}$
$∴$ Option (a) is correct.

AIEEE-2004

Complex Numbers and Quadratic Equation

118133 If $1, 2, 3$ and 4 are the roots of the equation $x^{4} +$ $a x + {b x}^{2} + c x + d = 0$ , then $a + 2 b + c$ is equal to

1 -25

2 0

3 10

4 24

Explanation:

C Given the equation
$x^{4} + a x^{3} + b x^{2} + cx + d = 0$
If $1, 2, 3$ and 4 are the roots of above
Then, $1 + a + b + c + d = 0$
$16 + 8 a + 4 b + 2 c + d = 0$
$81 + 27 a + a b + 3 c + d = 0$
$256 + 64 a 16 b + 4 c + d = 0$
Subtracting (i) from (ii) we get
$15 + 7 a + 3 b + c = 0$
$Subtracting (ii) from (iii)$
$65 + 19 a + 5 b + c = 0$
Subtracting (iii) from (iv)
$175 + 37 a + 7 b + c = 0$
Subtracting (v) from (vi), we get
$50 + 12 a + 2 b = 0$
Subtracting (vi) from (vii), we get
$110 + 18 a + 2 b = 0$
Subtracting (viii) and (ix), we get
$60 + 6 a = 0 \Rightarrow a = - 10$
$2 b = - 110 - 18 (- 10)$
$or 2 b = 70$
$∴ b = 35$
$∴ From eqn (v)$
$15 + 7 x (- 10) + 3 \times 35 + c = 0$
$or 15 - 70 + 105 + c = 0$
$or c = - 105 - 15 + 70$
$= - 120 + 70$
$= - 50$
$∴ a + 2 b + c = - 10 + 70 - 50$
$= 10$ $∴$ Option (c) is correct.

AP EAMCET-2007

Complex Numbers and Quadratic Equation

118130 If $α$ and $β$ are the roots of $x^{2} - 2 x + 4 = 0$ , then the value of $α^{6} + β^{6}$ is

1 32

2 64

3 128

4 256

Explanation:

C $α^{6} + β^{6} = {(α^{2})}^{3} + {(β^{2})}^{3}$
$= (α^{2} + β^{2}) ({(α^{2} + β^{2})}^{2} - 3 (α β)^{2})$
$We have the equation$
$x^{2} - 2 x + 4 = 0 and α, β are the roots of the eqn.$
$∴ α + β = 2$
$α β = 4$
$(α + β)^{2} = α^{2} + β^{2} + 2 α β$
$∴ α^{2} + β^{2} = (α + β)^{2} - 2 α β = 4 - 8 = - 4$
$∴ α^{6} + β^{6} = - 4 ((- 4)^{2} - 3 (4^{2})]$
$= - 4 (16 - 48)$
$= - 4 \times (- 32)$
$= 128$
$∴ Option (c) is correct.$

AP EAMCET-2009

Complex Numbers and Quadratic Equation

118131 If $α$ and $β$ are the roots of the equation $a x^{2} + b x$ $+ c = 0$ and, if ${p x}^{2} + q x + r = 0$ has roots $\frac{1 - α}{α}$ and $\frac{1 - β}{β}$ , then $r$ is equal to

1

a + 2 b

2

a + b + c

3

ab + bc + ca

4 abc

Explanation:

B $α and β are the roots of$
$a x^{2} + b x + c = 0$
$α + β = - b / a$
$α β = c / a$
$Given, {px}^{2} + qx + r = 0 has roots$
$\frac{1 - α}{α} and \frac{1 - β}{β}$
$\Rightarrow \frac{1 - α}{α} + \frac{1 - β}{β} = \frac{- q}{p}$
$and (\frac{1 - α}{α}) (\frac{1 - β}{β}) = \frac{r}{p}$
$or \frac{1}{α} + \frac{1}{β} = 2 - \frac{q}{p}$
$∴ \frac{(α + β)}{α β} = 2 - \frac{q}{p} = \frac{- b / a}{c / a} = \frac{- b}{c}$
$\frac{(1 - α) (1 - β)}{α β} = \frac{1 - α - β + α β}{α β}$
$= \frac{1 - (α + β) + α β}{α β}$
$= \frac{1 + \frac{b}{a} + \frac{c}{a}}{\frac{c}{a}} = \frac{a + b + c}{\frac{a}{c}} = \frac{a + b + c}{c}$
$\frac{r}{p} = \frac{a + b + c}{c}$
$r = a + b + c$
From eq $^{n}$ (i) $\frac{1}{α} - 1 + \frac{1}{β} - 1 = \frac{- q}{p}$
On comparing both side

AP EAMCET-2007

Complex Numbers and Quadratic Equation

118132 The set of values of $x$ for which the inequalities $x^{2} - 3 x - 10 < 0, 10 x - x^{2} - 16 > 0$ hold simultaneously, is

1

(- 2, 5)

2

(2, 8)

3

(- 2, 8)

4

(2, 5)

Explanation:

D Given $x^{2} - 3 x - 10 < 0 and$
$10 x - x^{2} - 16 > 0 hold.$
$∴ x^{2} - 3 x - 10 = x^{2} - 5 x + 2 x - 10$
$= x (x - 5) + 2 (x - 5)$
$= (x + 2) (x - 5) < 0$
$\Rightarrow - 2 < x < 5$
$Similarly, 10 x - x^{2} - 16$
$= - x^{2} + 10 x - 16 > 0$
$\Rightarrow x^{2} - 10 x + 16 < 0$
$\Rightarrow x^{2} - 8 x - 2 x + 16 < 0$
$\Rightarrow x (x - 8) - 2 (x - 8) < 0$
$\Rightarrow (x - 8) (x - 2) < 0$
$\Rightarrow 2 < x < 8$
$Both hold simultaneously ∴ 2 < x < 5$
$∴ x \in (2, 5)$
$∴ Option (d) is correct.$ Both hold simultaneously $∴ 2 < x < 5$

AP EAMCET-2007

Complex Numbers and Quadratic Equation

118164 If one root of the equation $x^{2} + p x + 12 = 0$ is 4 while the equation $x^{2} + p x + q = 0$ has equal roots, then the value of $q$ is

1

\frac{49}{4}

2 12

3 3

4 4

Explanation:

A Given one root of the equation
$x^{2} + px + 12 = 0$ is 4
$∴ (4)^{2} + 4 p + 12 = 0$
$\Rightarrow 4 p = - 12 - 16 = - 28$
$∴ p = \frac{- 28}{4} = - 7$
Also, given the equation $x^{2} + px + q = 0$ has equal roots $\Rightarrow D = 0, (p$ is criminate $) = 0$
i.e. $p^{2} - 4 q = 0$
$or (- 7)^{2} = 4 q$
$∴ q = \frac{49}{4}$
$∴$ Option (a) is correct.

AIEEE-2004

Complex Numbers and Quadratic Equation

118133 If $1, 2, 3$ and 4 are the roots of the equation $x^{4} +$ $a x + {b x}^{2} + c x + d = 0$ , then $a + 2 b + c$ is equal to

1 -25

2 0

3 10

4 24

Explanation:

C Given the equation
$x^{4} + a x^{3} + b x^{2} + cx + d = 0$
If $1, 2, 3$ and 4 are the roots of above
Then, $1 + a + b + c + d = 0$
$16 + 8 a + 4 b + 2 c + d = 0$
$81 + 27 a + a b + 3 c + d = 0$
$256 + 64 a 16 b + 4 c + d = 0$
Subtracting (i) from (ii) we get
$15 + 7 a + 3 b + c = 0$
$Subtracting (ii) from (iii)$
$65 + 19 a + 5 b + c = 0$
Subtracting (iii) from (iv)
$175 + 37 a + 7 b + c = 0$
Subtracting (v) from (vi), we get
$50 + 12 a + 2 b = 0$
Subtracting (vi) from (vii), we get
$110 + 18 a + 2 b = 0$
Subtracting (viii) and (ix), we get
$60 + 6 a = 0 \Rightarrow a = - 10$
$2 b = - 110 - 18 (- 10)$
$or 2 b = 70$
$∴ b = 35$
$∴ From eqn (v)$
$15 + 7 x (- 10) + 3 \times 35 + c = 0$
$or 15 - 70 + 105 + c = 0$
$or c = - 105 - 15 + 70$
$= - 120 + 70$
$= - 50$
$∴ a + 2 b + c = - 10 + 70 - 50$
$= 10$ $∴$ Option (c) is correct.

AP EAMCET-2007