Theory of Equations 1 Question 7
8. If one real root of the quadratic equation $81 x^{2}+k x+256=0$ is cube of the other root, then a value of $k$ is
(2019 Main, 11 Jan I)
(a) 100
(b) 144
(c) -81
(d) -300
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Solution:
- Given quadratic equation is
$$ 81 x^{2}+k x+256=0 $$
Let one root be $\alpha$, then other is $\alpha^{3}$.
Now, $\alpha+\alpha^{3}=-\frac{k}{81}$ and $\alpha \cdot \alpha^{3}=\frac{256}{81}$
$\left[\because\right.$ for $a x^{2}+b x+c=0$, sum of roots $=-\frac{b}{a}$
and product of roots $=\frac{c}{a}$ ]
$$ \begin{aligned} & \Rightarrow \quad \alpha^{4}=\frac{4}{3}^{4} \Rightarrow \alpha= \pm \frac{4}{3} \\ & \therefore \quad k=-81\left(\alpha+\alpha^{3}\right) \\ & =-81 \alpha\left(1+\alpha^{2}\right) \\ & =-81 \quad \pm \frac{4}{3} \quad 1+\frac{16}{9}= \pm 300 \end{aligned} $$
