polynomial of fifth degree Prove that the largest number of real roots of the equation $ x^5+a_1x^4+a_2x^3+a_3x^2+a_4x+a_5=0$ whose coefficients are real,is three if $2a_1^2-5a_2<0.$
My attempt is:
As coefficients are real,so complex roots will come in pair.Either one or three or five real roots are possible.I does not know what is relation between number of real roots and coefficients.Can someone guide me?I am cofused.
 A: Assuming that all the roots $\xi_1,\xi_2,\xi_3,\xi_4,\xi_5$ of your polynomial are real,
$$ (\xi_1+\xi_2+\xi_3+\xi_4+\xi_5)^2 \leq 5(\xi_1^2+\xi_2^2+\xi_3^2+\xi_4^2+\xi_5^2) \tag{1}$$
must hold: it is the Cauchy-Schwarz inequality. On the other hand, by Viète's theorem the LHS of $(1)$ is $a_1^2$, while the RHS of $(1)$ is $5(a_1^2-2a_2)$. So if all the roots of the polynomial are real,
$$ 2a_1^2 \geq 5a_2 \tag{2}$$
must hold. We have that $(2)$ does not hold, hence the real roots of our quintic polynomial are at most $3$ (since complex roots come in conjugated pairs).
A: Since non-real roots occur in conjugate pairs, number of real roots can either be $1$, $3$ or $5$. Assume, on the contrary, that all roots are real, i.e, $5$ real roots.   
Let $\{\alpha_{i}\}_{i=1}^5$ be the roots of the equation.   
By Vieta's Formula, 
$ \displaystyle -a_1= \sum_{i=1}^5 \alpha_{i} $ and $a_2 = \displaystyle \sum_{i < j} \alpha_{i}\alpha_{j}$   
Now, using the given condition, we have,   
$2\left(\displaystyle\sum_{i=1}^5 \alpha_{i}\right)^2 - 5\displaystyle\sum_{i < j} \alpha_{i}\alpha_{j} < 0$  
$\implies 2\displaystyle \sum_{i=1}^5 \alpha^2_{i} < \displaystyle\sum_{i < j} \alpha_{i}\alpha_{j}$  
However, by power-mean inequality $$\displaystyle \frac{1}{5} \sum_{i=1}^5 \alpha^2_{i} \geq \left(\displaystyle \frac{1}{5} \sum_{i=1}^5 \alpha_{i}\right)^2$$  
$\implies 2\displaystyle \sum_{i=1}^5 \alpha^2_{i} \geq \displaystyle\sum_{i < j} \alpha_{i}\alpha_{j}$  
Contradiction.
$\therefore$ The largest number of real roots is $3$
A: Apply the Tschirnhaus transformation $x=y-\frac{a_1}{5}$, which will change your equation to $$y^5+py^3+qy^2+ry+s=0$$
Your condition is precisely that $p>0$.  We now apply Descartes' rule of signs to the various cases of the signs of $q,r,s$.  For positive roots, we have the pattern $+,q,r,s$.  For negative roots, we have the pattern $-,q,-r,s$.  Exactly one of $+,q$ and $-,q$ has a sign change.  Exactly one of $q,-r$ and $q,r$ has a sign change.  Exactly one of $-r,s$ and $r,s$ has a sign change.  Hence there are just three real roots (provided $q,r,s$ are all nonzero).  I leave the cases of zero among $q,r,s$ for you.
A: A different approach from the ones presented above, using calculus. 
Assume that a polynomial of this form has 5 real roots. If the five are all distinct, then the mean-value theorem guarantees a zero of the derivative in each interval between the roots. That makes the derivative a quartic polynomial with four (distinct) real roots. Take two more derivatives, which leaves a quadratic with two (again distinct) real roots. This third derivative is $60x^2 + 24 a_1 x + 6 a_2$. For it to have two real roots, its discriminant must be positive, hence $24^2 a_1^2-4(60)(6)a_2 > 0$, or, equivalently, $2 a_1^2 - 5a_2 > 0$, which violates the condition $2a_1^2 - 5a_2 < 0$.
If there are four distinct real roots, then one of them is a double root. The mean-value theorem still guarantees a root of the derivative in each of the three intervals; there is also a root of the derivative at the same point as the double root of the original polynomial. That leaves four distinct roots for the first derivative, and the rest follows as before.
Thus, the assumption that there are more than 3 real roots leads to a contradiction, so there must be three or fewer real roots. Q. E. D.
A: Let $$p(x)=x^5+a_1x^4+a_2x^3+a_3x^2+a_4x+a_5$$ First Let me Explain for $p(x)$ having only one Real root, which means it should be Monotone Increasing(No Local Max or Local Minima).That is it's Derivative $p'(x)$ must not have any Real roots since $p'(x)$ should be above X axis.  That is
$$p'(x)=5x^4+4a_1x^3+3a_2x^2+2a_3x+a_4 \gt 0$$ and since the graph of $p'(x)$ Obvioulsly should have Local Minima(Since its graph resembles the graph of Vertical parabola), Then $p''(x)$ should have Exactly one Real root which means
$p''(x)$ should be Monotone Increasing and Hence $p'''(x) \gt 0$ That is
$$p'''(x)=60x^2+24a_1x+6a_2 \gt 0$$ $\implies$
$$10x^2+4a_1x+a_2 \gt 0$$ and from theory of Quadratic Equations if $ax^2+bx+c$ and $a$ have same sign then its Discriminant must be Negative.
Hence
$$16a_1^2-40a_2 \lt 0$$ $\implies$ 
$$2a_1^2-5a_2 \lt 0$$. The same analysis can be applied for $p(x)$ having three real roots.
A: Hint: Use Vieta's theorem to obtain $2(\sum_i x_i)^2<5\sum_{i<j} x_ix_j$. Can this be possible if all the roots $x_i$'s are real?
