Question

If $\alpha ,\beta ,\gamma$ are the roots of equation $x^3-6x^2+11x-6=0$, then find the equation whose roots are ${\alpha }^2+{\beta }^2,{\beta }^2+{\gamma }^2,{\gamma }^2+{\alpha }^2$ 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

The correct option is A $x^3-28x^2+245x-650=0$

Given equation: $x^3-6x^2+11x-6=0$
Clearly, $x=1$ is a root of the equation.(By trail and error)
Factorising the equation,
$x^3-6x^2+11x-6=0\Rightarrow (x-1)(x^2-5x+6)=0\Rightarrow (x-1)(x-2)(x-3)=0$
Therefore, the roots of the equation are $1,2,3$
Let $\alpha =1,\beta =2,\gamma =3$
Now,
${\alpha }^2+{\beta }^2=5{\beta }^2+{\gamma }^2=13{\alpha }^2+{\gamma }^2=10$
So, the required equation is,
$(x-5)(x-13)(x-10)=0\Rightarrow (x-5)(x^2-23x+130)=0$

Hence, the required equation is
$x^3-28x^2+245x-650=0$