Question

If $\alpha$ and $\beta$ 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 $\beta =-4\alpha$
• C. $\alpha =3\beta$ or $\beta =3\alpha$
• D. $\alpha =-3\beta$ or $\beta =-3\alpha$

Answer 1

Answer: (C)

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

Given that $(\alpha ,\beta )$ are roots of the quadratic equation
$a{x}^2+bx+c=0$
and $3b^2=16ac.....\left(i\right)$

$\therefore$ $\alpha +\beta =\frac{-b}{a}$ and $\alpha \beta =\frac{c}{a}.....\left(ii\right)$

From eq $\left(i\right)$ we get
$2b.b=16a.c$

$\Rightarrow$ $\Rightarrow 3{\left(\frac{b}{a}\right)}^2=16\left(\frac{c}{a}\right)$ (divide by $a^2$)$\Rightarrow 3{(\alpha +\beta )}^2=16\alpha \beta$

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

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

$\Rightarrow 3\left(\frac{\alpha }{\beta }\right)+3\left(\frac{\beta }{\alpha }\right)=10$

Let $\frac{\alpha }{\beta }=x$

Then, $3x+\frac{3}{x}=10$

$\Rightarrow 3x^2-10x+3=0$

$\Rightarrow 3x^2-9x-x+3=0$

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

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

$\Rightarrow x=\frac{1}{3},3$ $\Rightarrow$ $\frac{\alpha }{\beta }=\frac{1}{3},3$

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