Question

If the cubic $y=x^3+px+q$ has $3$ distinct real roots, then $4p^3+27q^2$ is

• A. less than $0$
• B. greater than $0$
• C. equal to $0$
• D. none of these

Answer 1

The correct option is A less than $0$

Now, since all three roots are distinct

Let $x_1,x_2$ be the points of maxima and minima of the equation $y=f\left(x\right)=x^3+px+q$
$\therefore x_1,x_{2}aretherootsoftheequation{f}^{\prime }\left(x\right)=3x^2+p$
$f^{\prime }\left(x\right)=3x^2+0.x+p$
So, $x_1+x_2=0$ and $x_1{x}_2=\frac{p}{3}$

$f\left(x_1\right)\times f\left(x_2\right)<0$
$\Rightarrow (x_1^3+p{x}_1+q)(x_2^3+p{x}_2+q)<0$

$\Rightarrow x_1^3.x_2^3+p{x}_1^3{x}_2+q{x}_1^3+p^2{x}_1{x}_2+p{x}_1{x}_2^3+q{x}_2^3+pq{x}_1+q^2+pq{x}_2<0$

$\Rightarrow {\left(x_1{x}_2\right)}^3+p{x}_1{x}_2(x_1^2+x_2^2)+q(x_1^3+x_2^3)+pq(x_1+x_2)+p^2{x}_1{x}_2+q^2<0$

$\Rightarrow {\left(x_1{x}_2\right)}^3+p{x}_1{x}_2\{{(x_1+x_2)}^2-2x_1{x}_2\}+q\{{(x_1+x_2)}^3-3(x_1{x}_2\left)\right(x_1+x_2\left)\right\}+pq(x_1+x_2)+p^2{x}_1{x}_2+q^2<0$

$\Rightarrow \frac{p^3}{27}+\frac{p^2}{3}\{-\frac{2p}{3}\}+p^2\left\{\frac{p}{3}\right\}+q^2<0$

$\Rightarrow 4p^3+27q^2<0$