Question

If one of the roots of the equation $x^2+rx-s=0$ is the square of other, then $r^3+s^2+3sr-s=$

• A. $0$
• B. $s$
• C. $1$
• D. $r$

Answer 1

The correct option is A $0$

Let, $\alpha$ be the one root of equation, $x^2+rx-s=0$, then ${\alpha }^2$ will be other root of $x^2+rx-s=0$.

$\therefore \alpha +{\alpha }^2=-r\cdots \left(i\right)\&\alpha .{\alpha }^2=-s\Rightarrow {\alpha }^3=-s$

Now, $\left[\alpha \right(1+\alpha ){]}^3=(-r{)}^3$

$\Rightarrow {\alpha }^3(1+{\alpha }^3+3\alpha (1+\alpha \left)\right)=-r^3\cdots \left(ii\right)$

$\Rightarrow (-s)\left[\right(1-s)+3(-r\left)\right]=-r^3$

$\Rightarrow s^2-s+3sr=-r^3$

$\Rightarrow r^3+s^2+3sr-s=0$