Question

If $\alpha , \beta , \gamma$ are the roots of $x^3 - 6x^2 + 11x - 6 = 0 $ , then the equation having the roots $\alpha^2 + \beta^2 + \gamma^2$ and $\gamma^2 + \alpha^2$ is

• A. $x^3 - 28x^2 + 245x - 650 = 0$
• B. $x^3 - 28x^2 + 245 x + 650 = 0 $
• C. $x^3 + 28x^2 - 245x - 650 =0 $
• D. $x^3 + 28x^2 + 245x - 650 = 0 $

Answer 1

The Correct Option is A

$x^{3}-6 x^{2}+11 x-6=0 \dots$(i)
$\Rightarrow(x-1)(x-2)(x-3)=0$
$\Rightarrow x=1,2,3$
$\because \alpha, \beta, \gamma$ are the roots of the E(i), so
$\alpha=1, \beta=2, \gamma=3$
Therefore, $\alpha^{2}+\beta^{2}=(1)^{2}+(2)^{2}=5=\alpha'$ (say)
$\beta^{2}+\gamma^{2}=(2)^{2}+(3)^{2}=13=\beta'($ say $)$
and $\gamma^{2}+\alpha^{2}=(3)^{2}+1=10=\gamma'($ say $)$
Equation of the having the roots $\alpha', \beta'$ and $\gamma'$ ,
$x^{3}-\left(\alpha'+\beta'+\gamma'\right) x^{2}+\left(\alpha' \beta'+\beta'\gamma'+\gamma' \alpha'\right) x$
$-\alpha' \beta' \gamma'=0 $
$\Rightarrow x^{3}-(5+13+10) x^{2}+(5 \times 13+13 \times 10+10 \times 5) x$
$-5 \times 13 \times 10=0$
$\Rightarrow x^{3}-28 x^{2}+245 x-650=0$