Question

The equation $x^2+px-q=0$ has one root as a square root of the other.
If $p^3+q^2=q(1+kp)$, find the value of k.

• A. 4
• B. -1
• C. -2
• D. -3

Answer 1

The correct option is D

-3

Let the roots of the equation be $\alpha ,{\alpha }^2$.
$\alpha +{\alpha }^2=-p$ -(i)
${\alpha }^3=-q$-(ii)
Cubing (i) gives
$(\alpha +{\alpha }^2{)}^3=[\alpha (\alpha +1){]}^3$
$-p^3={\alpha }^3({\alpha }^3+3\alpha (\alpha +1)+1)$
$={\alpha }^3({\alpha }^3+3(\alpha +{\alpha }^2)+1)$
$=-q(-q+3(-p)+1)$
$-p^3=q^2+3pq-q$
$p^3+q^2=q-3pq$
$=q(1-3p)$
$p^3+q^3=q(1+kp)$ (given)
$\Rightarrow k=-3$