Question

If $\alpha$ is a real root of the equation $x^3+p{x}^2+qx+r=0$, where p, q and r are real. If $p^2-4q-2p\alpha -3{\alpha }^2\ge 0$ then other roots are ________.

• A. Imaginary conjugate
• B. Irrational conjugate
• C. Real numbers
• D. None of these

Answer 1

The correct option is C

Real numbers

$\alpha$ is a one root of the equation $x^3+p{x}^2+2x+r=0$ __________(1)

It should satisfy the equation.

${\alpha }^3+p{\alpha }^2+q\alpha +r=0$

Also, $(x-\alpha )$ is a factor of $x^3+p{x}^2+qx+r.$
So, divide this expression by $(x-\alpha )$, we get

$\begin{array}{cc}& x^2+(\alpha +p)x+({\alpha }^2+p\alpha +q)\\ x-\alpha & x^3+p{x}^2+qx+r\\ & \_x^3\_+x^2\alpha \\ & (p+\alpha )x^2+qx\\ & {}_{-}(p+\alpha )x^2{}_{+}-\alpha (\alpha +p)x\\ & ({\alpha }^2+p\alpha +q)x+r\\ & {}_{-}({\alpha }^2+p\alpha +q)x{}_{+}-{\alpha }^3+p{\alpha }^2+q\alpha \\ & {\alpha }^3+p{\alpha }^2+q\alpha +r\end{array}$
Given ${\alpha }^3+p{\alpha }^2+q\alpha +r=0$

Then cubic equation can be written as

$(x-\alpha )(x^2+(\alpha +p)x+({\alpha }^2+p\alpha +q\left)\right)=0$

Now, $x^2+(\alpha +p)x+({\alpha }^2+p\alpha +q)=0$,
$D=(\alpha +p{)}^2-4(1\left)\right({\alpha }^2+p\alpha +q)$
$=p^2-4q-2p\alpha -3{\alpha }^2\ge 0[\because \text{Given}]\Rightarrow D\ge 0\text{for}{x}^2+(\alpha +p)x+({\alpha }^2+p\alpha +q)=0$

So, other two roots of the given cubic equation are real.