Q100 / 200 · WBJEE 2013 · Mathematics
If $\alpha$ and $\beta$ are the roots of the quadratic equation $a x^{2}+b x+c=0$ and $3 b^{2}=16 a c$ then
- A$\alpha=4 \beta$ or $\beta=4 \alpha$
- B$\alpha=-4 \beta$ or $\beta=-4 \alpha$
- Correct answerC$\alpha=3 \beta$ or $\beta=3 \alpha$
- D$\alpha=-3 \beta$ or $\beta=-3 \alpha$
Solution
Given that, $(\alpha, \beta)$ are roots of the quadratic
equation
and $\begin{aligned} a x^{2}+b x+c &=0 \\ 3 b^{2} &=16 a c \end{aligned}$
$\therefore$
$\alpha+\beta=\frac{-b}{a}$ and $\alpha \beta=\frac{c}{a}$
From Eq. (i), we get $\quad 3 b \cdot b=16 \cdot a \cdot c$
$\Rightarrow \quad 3\left(\frac{b}{a}\right)^{2}=16\left(\frac{c}{a}\right)$ (divide by $\left.a^{2}\right)$
$\Rightarrow \quad 3(\alpha+\beta)^{2}=16 \alpha \beta[$ from Eq. $(i i)]$
$\Rightarrow 3 \alpha^{2}+3 \beta^{2}+6 \alpha \beta=16 \alpha \beta$
$\Rightarrow \quad 3 \alpha^{2}+3 \beta^{2}=10 \alpha \beta$
$\Rightarrow \quad 3\left(\frac{\alpha}{\beta}\right)+3\left(\frac{\beta}{\alpha}\right)=10$
Let $\quad \frac{\alpha}{\beta}=x$
$\begin{array}{lr}\text { Then, } & 3 x+\frac{3}{x}=10 \\ \Rightarrow & 3 x^{2}-10 x+3=0 \\ \Rightarrow & 3 x^{2}-9 x-x+3=0 \\ \Rightarrow & 3 x(x-3)-1(x-3)=0 \\ \Rightarrow & (x-3)(3 x-1)=0 \\ \Rightarrow & x=\frac{1}{3}, 3 \Rightarrow \frac{\alpha}{\beta}=\frac{1}{3}, 3\end{array}$
$\therefore \quad \beta=3 \alpha$ or $\alpha=3 \beta$
Hence, the correct answer is option: 'C'.