Question

If $a,b,c,d,e$ are real numbers,show that the roots of $x^5+a{x}^4+b{x}^3+c{x}^2+dx+e=0$ cannot all be real, if $2a^2<5b$.

Answer 1

Here, $x^5+a{x}^4+b{x}^3+c{x}^2+dx+e=0$ has real roots ${\alpha }_1$, ${\alpha }_2$, ${\alpha }_3$, ${\alpha }_4$, ${\alpha }_5$.
$\Rightarrow$ ${\alpha }_1+{\alpha }_2+{\alpha }_3+{\alpha }_4+{\alpha }_5=\sum {\alpha }_i=-a$
${\sum }_{i<j}{\alpha }_i{\alpha }_j=b$ (i)
${(\sum {\alpha }_i)}^2=a^2$ (ii)
$\Rightarrow$ $\sum {\alpha }_i^2$+2\left ( \sum_{i< j}\alpha _{i}\alpha _{j} \right )=a^{2}$\left(iii\right)$\Rightarrow \left ( \sum \alpha _{i}^{2} \right )=a^{2}-2b$[fromEqs.(i),(ii\left)and\right(iii\left)\right]Asweknow,$\displaystyle \left ( \frac{\alpha _{1}^{2}+\alpha _{2}^{2}+\alpha _{3}^{2}+\alpha _{4}^{2}+\alpha _{5}^{2}}{5} \right )\geq \left ( \frac{\alpha _{1}+\alpha _{2}+\alpha _{3}+\alpha _{4}+\alpha _{5}}{5} \right )^{2}$\left(knownasTchebychef{f}^{\prime }sinequality\right)$\Rightarrow 5\left ( a^{2}-2b \right )\geq a^{2}$or$4a^{2}\geq 10b$or$2a^{2}\geq 5b$\left(forrealroots\right)Thus,if$2a^{2}< 5b$ cannot all be real roots.