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

Answer 1

Answer: (A)

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