Question

If $\alpha$ be 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. Real numbers
• B. Imaginary conjugate
• C. Irrational conjugate
• D. None of these

Answer 1

The correct option is A

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. Divide this expression by $x-\alpha$.

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

Then cubic equation can be written as

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

$I{t}^{\prime }sgiven{p}^2-4q-2p\alpha -3{\alpha }^2\ge 0$

We will get it only when D $\ge$ 0

$b^2$ - 4ac $\ge$ 0

${(\alpha +p)}^2-4({\alpha }^2+p\alpha +q)\ge 0$

$p^2-4q-2p\alpha -3{\alpha }^2\ge 0$

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