Question

Find the value of a if one root of the equation $8x^2-6x+a=0$ is the square of the other.

Answer 1

$8x^2-6x+a=0$

Let root be $\alpha ,{\alpha }^2$

$\therefore \alpha +{\alpha }^2=\frac{6}{8}$

$2{\alpha }^2=\frac{9}{8}$

$2(1+\alpha )=\frac{6}{8}$

${\alpha }^3=\frac{9}{8}$

$\alpha =\frac{6}{8},\alpha =1-\frac{6}{8}=\frac{1}{4}$

If, $\alpha =\frac{6}{8}$

$\Rightarrow a=\frac{6\times 6\times 6\times 8}{8\times 8\times 8}=\frac{27}{8}$

And, $\alpha =\frac{1}{4}$

$\Rightarrow a=\frac{1\times 8}{4\times 4\times 4}=\frac{1}{8}$