Question

If one root of the equation $8x^2-6x+k=0$ is the square of the other, then the value of $k$ is

• A. $0,3$
• B. $-1,27$
• C. $0,-2$
• D. $1,-27$

Answer 1

The correct option is D $1,-27$

Let one root of the equation be $\alpha$.
Then the other root is ${\alpha }^2$.
Now sum of roots is
$=\frac{-b}{a}$
$=\frac{-(-6)}{8}$
$=\frac{3}{4}$
Hence
$\alpha +{\alpha }^2=\frac{3}{4}$
Or
${\alpha }^2+\alpha -\frac{3}{4}=0$
$(\alpha +\frac{1}{2}{)}^2-\frac{1}{4}-\frac{3}{4}=0$
$(\alpha +\frac{1}{2}{)}^2=1$
$\alpha +\frac{1}{2}=\pm 1$
Hence
$\alpha =\frac{1}{2}$ and $\alpha =\frac{-3}{2}$
Or
${\alpha }^2=\frac{1}{4}$ and ${\alpha }^2=\frac{9}{4}$
Hence product of roots is
${\alpha }^3=\frac{1}{2}\times \frac{1}{4}$
$=\frac{1}{8}$
And
${\alpha }^3=(\frac{-3}{2}{)}^2$
$=\frac{-27}{8}$
Hence product of roots
$=\frac{c}{a}$
$=\frac{k}{8}$
$=\frac{1}{8}$ or $\frac{-27}{8}$
Hence
$k=1,-27$.