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Prove that if $2a^2<15b$ , not all roots of $x^5 - ax^4 + 3bx^3 + cx^2 + dx + e = 0$ can be real. It is given that $a,b,c,d,e$ belongs to Real number.

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    This doesn't seem correct, don't all odd degree real coefficients polynomials have a real root by Intermediate Value Theorem? – 2017-02-12
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    I think you mean "not all roots can be real". – 2017-02-12

2 Answers 2

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Let's call $p(x)=x^5-ax^4+3bx^3+cx^2+dx+e$ and its roots $x_1, x_2, x_3, x_4, x_5$. Vieta's formulas gives us

$$ a= x_1+x_2+x_3+x_4+x_5$$

and

$$ 3b = x_1x_2+x_1x_3+x_1x_4+x_1x_5+x_2x_3+x_2x_4+x_2x_5+x_3x_4+x_3x_5+x_4x_5$$

If $x_1, x_2, x_3, x_4, x_5 \in \mathbb{R}$, we can use Rearrangement Inequallity to get

$$ x_1x_2+x_1x_3+x_1x_4+x_1x_5+x_2x_3+x_2x_4+x_2x_5+x_3x_4+x_3x_5+x_4x_5 \leq 2(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2) $$

Now, it's easy to see that

$$ \begin{align*} & 2a^2-15b \\ =& 2(x_1+x_2+x_3+x_4+x_5)^2-5(x_1x_2+x_1x_3+x_1x_4+x_1x_5+x_2x_3+x_2x_4+x_2x_5+x_3x_4+x_3x_5+x_4x_5) \\ =& 2(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2)-(x_1x_2+x_1x_3+x_1x_4+x_1x_5+x_2x_3+x_2x_4+x_2x_5+x_3x_4+x_3x_5+x_4x_5) \\ \geq&\, 0 \end{align*}$$

So, if all roots are real, we get $2a^2 \geq 15b$. Now, if $2a^2<15b$, we can't have all roots real.

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Hint: in general, if a polynomial $P(x)$ has $n$ real roots (counting multiplicities) then $P'(x)$ has at least $n-1$ real roots (also counting multiplicities), then $P''(x)$ has at least $n-2$ real roots etc.

For $P(x) = x^5 - ax^4 + 3bx^3 + cx^2 + dx + e$ to have all the roots real, it is therefore necessary that $P'''(x)= 6\,(10x^2-4a x+3 b)$ has all its roots real. But $P'''(x)$ is a quadratic, and its roots are real iff the discriminant $\Delta=36\cdot 4\,(4a^2-30b) \ge 0 \;\iff\; 2a^2 \ge 15b\,$.