Question

If one root of the equation $x^2+px+q=0$ is the square of the other, then

• A. $p^3+q^2-q(3p+1)=0$
• B. $p^3+q^2+q(1+3p)=0$
• C. $p^3+q^2+q(3p-1)=0$
• D. $p^3+q^2+q(1-3p)=0$

Answer 1

Answer: (D)

$x^2+px+q=0$
Let, $\alpha ,{\alpha }^2$ are the roots of the given equation.
$\Rightarrow \alpha +{\alpha }^2=-p$
and $\alpha \cdot {\alpha }^2=q$
$\Rightarrow {\alpha }^3=q$
$\because \alpha +{\alpha }^2=-p$
$\Rightarrow {\left(\alpha +{\alpha }^2\right)}^3=-p^3$
$\Rightarrow {\alpha }^6+{\alpha }^3+3\alpha \cdot {\alpha }^2\left(\alpha +{\alpha }^2\right)=-p^3$
$\Rightarrow q^2+q+3{\alpha }^3\left(\alpha +{\alpha }^2\right)=-p^3$
$\Rightarrow q^2+q+3q(-p)=-p^3$
$\Rightarrow p^3+q^2+q-3pq=0$
$\Rightarrow p^3+q^2+q(1-3p)=0$