Question

If $\alpha ,\beta$ are the roots of the equation $2x^2-3x-6=0$, then the equation whose roots are ${\alpha }^2-1$ and ${\beta }^2-1$ is

• A. $2x^2-25x-7=0$
• B. $2x^2-17x+7=0$
• C. $4x^2-17x+7=0$
• D. $4x^2-25x+7=0$

Answer 1

The correct option is D $4x^2-25x+7=0$

Let ${\alpha }^2-1=y\Rightarrow \alpha =\pm \sqrt{y+1}$

Now, $\alpha$ is the root of the equation $2x^2-3x-6=0$, so
$2{\alpha }^2-3\alpha -6=0$
$\Rightarrow 2(\pm \sqrt{y+1}{)}^2-3(\pm \sqrt{y+1})-6=0$
$\Rightarrow 2y+2-6=(\pm 3\sqrt{y+1})$
$\Rightarrow 2y-4=(\pm 3\sqrt{y+1})$
Squaring on both sides, we get
$\Rightarrow 4y^2-16y+16=9(y+1)$
$\Rightarrow 4y^2-25y+7=0$

Hence, the required equation is $4x^2-25x+7=0$