Question

If $\alpha ,\beta$ are the roots of the quadratic equation $a{x}^2+bx+c=0$ and $3b^2=16ac$ then

• A. $\alpha =4\beta$ or $\beta =4\alpha$
• B. $\alpha =-4\beta$ or $\beta =-\alpha$
• C. $\alpha =3\beta$ or $\beta =3\alpha$
• D. $\alpha =-3\beta$ or $\beta =-3\alpha$

Answer 1

Answer: (C)

$\alpha =3\beta$ or $\beta =3\alpha$

$\alpha +\beta =-b/a$

$\Rightarrow \alpha \beta =c/a$

$\Rightarrow (\alpha +\beta {)}^2=b^2/a^2$

Given $3b^2=16ac$

$\Rightarrow 3(\alpha +\beta {)}^2{a}^2=16.a.a(\alpha \beta )$

$\Rightarrow 3{\alpha }^2+3{\beta }^2+6\alpha \beta =16\alpha \beta$

$\Rightarrow 3{\alpha }^2-10\alpha \beta +3{\beta }^2=0$

$\Rightarrow (\alpha -3\beta )(3\alpha -\beta )=0$

$\Rightarrow \alpha =3\beta$ or $\Rightarrow \beta =3\alpha$