Question

If $\alpha$ and $\beta$ are the roots of the equation $2x^2-7x+8=0$, then the equation whose roots are $(3\alpha -4\beta$) and ($3\beta -4\alpha$) is

• A. $2x^2+7x+98=0$
• B. $x^2+7x+98=0$
• C. $2x^2-7x-98=0$
• D. $2x^2-7x+98=0$

Answer 1

The correct option is A $2x^2+7x+98=0$

$2x^2-7x+8=0$

$\alpha ,\beta$ are roots of above equation $(a{x}^2+bx+c=0)$

$\therefore \alpha +\beta =\frac{-b}{a}=\frac{7}{2}$ -(1)

$\alpha \beta =\frac{8}{2}\left(\frac{c}{a}\right)=4$

Roots of other $e{q}^n\Rightarrow (3\alpha -4\beta ),(3\beta -4\alpha )$

$\Rightarrow$ sum of roots = $3\alpha -4\beta +3\beta -4\alpha$

$=-\alpha -\beta =-(\alpha +\beta )$

$=(-\frac{7}{2})$ (from (1))

$\Rightarrow$ product of roots $\Rightarrow (3\alpha -4\beta )(3\beta -4\alpha )$

$=9\alpha \beta -12{\alpha }^2-12{\beta }^2+16\alpha \beta$

$=25\alpha \beta -12({\alpha }^2+{\beta }^2)$

$=25\alpha \beta -12\left(\right(\alpha +\beta {)}^2-2\alpha \beta )$

$=25\alpha \beta -12(\alpha +\beta {)}^2+21\alpha \beta$

$=49\alpha \beta -12{(\alpha +\beta )}^2$

$\Rightarrow 49\left(4\right)-12(\frac{7}{2}{)}^2$

$\Rightarrow 49\left(4\right)-12\times \frac{49}{4}=49(4-3)$

$\Rightarrow \left(49\right)$

$\Rightarrow$ quadratic poly $\Rightarrow x^2$ - (sum of roots)x+(product of roots)

$\Rightarrow x^2-\left(\frac{-7}{2}\right)x+49$

$\Rightarrow 2x^2+7x+98$ (option A)