Question

If $\alpha ,\beta ,\gamma$ are the roots of $x^3-7x+6=0$ then the equation whose roots are ${(\alpha +\beta )}^2,{(\beta +\gamma )}^2,{(\gamma +\alpha )}^2$ is

• A. $x(x^2+7)=36$
• B. $x{(x+7)}^2=36$
• C. $x(x^2-7)=36$
• D. $x{(x-7)}^2=36$

Answer 1

Answer: (C)

$x(x^2-7)=36$

$x^3-7x+6=0$

$\Rightarrow (x-1)(x+3)(x-2)=0$

$\therefore \alpha =-3,\beta =1,\gamma =2$

$(\alpha +\beta {)}^2=4,(\beta +\gamma {)}^2=9,(\gamma +\alpha {)}^2=1$

$\therefore (x-4)(x-9)(x-1)=0$

$\Rightarrow (x^2-9x-4x+36)(x-1)=0$

$\Rightarrow (x-1)(x^2-13x+36)=0$

$x^3-13x^2+36x-x^2+13x-36=0$

$\Rightarrow x^3-14x^2+49x-36=0$

$\Rightarrow x(x^2-14x+49)=36$

$\Rightarrow x(x-7{)}^2=36$