Question

If $\alpha$ and $\beta$ are the roots of the equation $2x^2-5x-7=0$, then the equation whose roots are $2\alpha +3\beta$ and $3\alpha +2\beta$ is ________.

• A. $2x^2+25x-68=0$
• B. $x^2+25x+68=0$
• C. $2x^2-25x-68=0$
• D. $2x^2-25x+68=0$

Answer 1

The correct option is D $2x^2-25x+68=0$

For the equation whose roots are $2\alpha +3\beta ,3\alpha +2\beta$,

The sum of the roots $=2\alpha +3\beta +3\alpha +2\beta =5(\alpha +\beta )\to \left(I\right)$
The product of the roots $=(2\alpha +3\beta )(3\alpha +2\beta )$
$=6{\alpha }^2+13\alpha \beta +6{\beta }^2$

$=6({\alpha }^2+{\beta }^2)+13\alpha \beta$

$=6\left(\right(\alpha +\beta {)}^2-2\alpha \beta )+13\alpha \beta$

$=6(\alpha +\beta {)}^2-12\alpha \beta +13\alpha \beta$

$=6(\alpha +\beta {)}^2+\alpha \beta$ $\to \left(II\right)$

We also know $\alpha$, $\beta$ are the roots of the equation $2x^2-5x-7=0$

$\Rightarrow \alpha +\beta =5/2$ $\to \left(III\right)$ and,

$\alpha \beta =-7/2$ $\to \left(IV\right)$

Substituting $\left(III\right),\left(IV\right)$ in $\left(I\right),\left(II\right)$ for the equation to be found we get,

Sum of the roots $=5\times \frac{5}{2}=\frac{25}{2}$
Product of the roots$=6\times (5/2{)}^2-7/2=\frac{150}{4}-\frac{7}{2}=\frac{136}{4}=\frac{68}{2}$
$\therefore$ the equation to be found is $x^2-\left(\text{Sum of roots}\right)x+\text{Product of roots}=0$

$\Rightarrow x^2-\left(25/2\right)x+68/2=0$

$\Rightarrow 2x^2-25x+68=0$