Question

If ${2x}^3+{ax}^2+bx+4=0$ (a and b are positive real numbers) has real roots,
then $a+b>6(p^{1/3}+q^{1/3})$
Find $|p-q|$

Answer 1

Let $\alpha ,\beta ,\gamma$ be the roots of the given equation. Then,
$\alpha +\beta +\gamma =-\frac{a}{2}........\left(i\right)\alpha \beta +\beta \gamma +\gamma \alpha =\frac{b}{2}........\left(ii\right)and\alpha \beta \gamma =-2.........\left(iii\right)$
Since all the coefficient of the given equation are positive, so its all roots are negative. Let $\alpha =-{\alpha }_1,\beta =-{\beta }_{1}and\gamma =-{\gamma }_1$
$\therefore$ from (i), (ii),and (iii) we obtain
${\alpha }_1+{\beta }_1+{\gamma }_1=a/2........\left(iv\right){\alpha }_1{\beta }_1+{\beta }_1{\gamma }_1+{\gamma }_1{\alpha }_1=b/2........\left(v\right)and{\alpha }_1{\beta }_1{\gamma }_1=2.........\left(vi\right)NowA.M.\ge G.M.\Rightarrow \frac{{\alpha }_1+{\beta }_1+{\gamma }_1}{3}{\ge \left({\alpha }_1{\beta }_1{\gamma }_1\right)}^{1/3}\Rightarrow \frac{a}{6}\ge 2^{1/3}\Rightarrow a\ge 6\left(2^{1/3}\right)........\left(vii\right)$
$AgainA.M.\ge G.M.\Rightarrow \frac{{\alpha }_1{\beta }_1+{\beta }_1{\gamma }_1+{\gamma }_1{\alpha }_1}{3}{\ge \left\{\left({\alpha }_1{\beta }_1\right)\left({\beta }_1{\gamma }_1\right)\left({\gamma }_1{\alpha }_1\right)\right\}}^{1/3}\Rightarrow b\ge 6\left(2^{2/3}\right)........\left(viii\right)$
from (vii) and (viii), we obtain
$a+b\ge 6(2^{1/3}+2^{2/3})\Rightarrow a+b>6(2^{1/3}+4^{1/3})\Rightarrow |p-q|=|2-4|\therefore |p-q|=2$