Question

If the two equations $x^3+3p{x}^2+3qx+r=0$ and $x^2+2px+q=0$ have a common root, then the value of $4(p^2-q)(q^2-pr)$ is

• A. $(pq+r{)}^2$
• B. $(pq-r{)}^2$
• C. $(p-qr{)}^2$
• D. $(p+qr{)}^2$

Answer 1

The correct option is B $(pq-r{)}^2$

Let $\alpha$ be the common root.
$\Rightarrow {\alpha }^3+3p{\alpha }^2+3q\alpha +r=0\cdots \left(1\right)$
and ${\alpha }^2+2p\alpha +q=0\cdots \left(2\right)$
$\left(1\right)-\alpha \left(2\right)$, we get
$p{\alpha }^2+2q\alpha +r=0\cdots \left(3\right)$

It can be observed that equation $\left(2\right)$ and $\left(3\right)$ have $\alpha$ as common root.
So by applying common root condition $(a_1{b}_2-a_2{b}_1)(b_1{c}_2-b_2{c}_1)=(a_1{c}_2-a_2{c}_1{)}^2$, we get
$(2q-2p^2)(2pr-2q^2)=(r-pq{)}^2$
$\Rightarrow 4(p^2-q)(q^2-pr)=(pq-r{)}^2$