# Real Roots and Differentiation

Prove that the equation $$x^5 − 1102x^4 − 2015x = 0$$ has at least three real roots.

So do I sub in values of negative and positive values of $$x$$ to show that there are at least three real roots? The method to do this question is not by finding the factors of $$x$$ right? Because it will be too tedious so I want to ask whats the other solution to prove this? Help appreciated. Thank you very much.

• one root is $x=0$. So your problem is reduced to $x^4-1102x^3-2015=0$. Then, try to find two pairs $x_1, x_2$ where $x^4-1102x^3-2015$ changes sign. You can then conclude from the Intermediate value theorem, that there are at least 2 additional roots. – MrYouMath Sep 17 '15 at 18:10
• let $$f(x)=x^5-1102x^4-2015x$$ then calculate $$f(-2)$$ and $$f(-1)$$ and $$f(1000)$$ and $$f(2000)$$ – Dr. Sonnhard Graubner Sep 17 '15 at 18:13

It's clear that $x=0$ is one of the roots. Hence, if we prove there are atleast 2 zeros to $f(x) := x^4-1102x^3-2015$, we are done.

Observe, $f(0) < 0$ and $f(-2) > 0$, so from Intermediate Value Theorem there exists at least one root between $-2$ and $0$.

Now, lets say there is exactly one real root to $f$ which means that there are 3 non real complex roots to $f$. This can not be possible as complex roots occur in conjugate pairs. Hence, there are at least 2 real roots to $f=0$

It's about the real zeros of $x q(x)$ with $q(x):=x^4-1102 x^3-2015$. There is the obvious zero $x=0$. Furthermore $q(0)<0$ and $\lim_{x\to\pm\infty} q(x)=+\infty$ guarantee two more real zeros.

You can use Descartes' rule of signs to tell you the number of real roots as long as you are not interested in the value of each.

First observe that $x=0$ is a root for:

$f(x)=x^5 − 1102x^4 − 2015x$

Second, count positive real roots by counting sign changes in $f(x)$, we have: (+-)(--), that is 1 sign change indicating 1 positive root.

third, count negative real roots by counting sign changes in $f(-x)$ where: $f(-x)=-x^5 - 1102x^4 + 2015x$

here we have the signs: (--)(-+), so we have 1 negative root.

From the above, we have 3 real roots for $f(x)$.

$$x^5 − 1102 \cdot x^4 − 2015 \cdot x = 0$$

This factors into,

$$\left( x^4 − 1102x^3 − 2015 \right) \cdot x = 0$$

Therefore $x=0$ is a root. Keep simplifying,

$$x^4 − 1102x^3 − 2015=0$$

Set this equal to $f(x)$,

$$f(x)=x^4 − 1102x^3 − 2015$$

$f(-1)=-912$ and $f(-2)=6817$, thus by the Intermediate Value Theorem, there is a zero in-between $-2$ and $-1$.

The same thing holds for $f(1100)$ and $f(1110)$

Now, in the spirit of fairness, let's quickly come up with the method that will allow us to "guess" where the roots of $f(x)$ lie. We'll use Newton's Method.

We get,

$$x_{n+1}=x_{n}-{{x^4 − 1102x^3 − 2015} \over {4 \cdot x^3-2204 \cdot x^2}}$$

Now, the property we'll use is the fact that the convergence for the method is oscillatory. That means if you guess too low, then the next guess will be to high $^1$. Applying this principle results in a guess of $1000$ for the root resulting in a new guess of $1170$ for the next root. Once again, the Intermediate Value Theorem applies.

$^1$ (There are subtleties about convergence and when this works and doesn't work, but generally speaking, this is the case if you pick a reasonable guess)

• must we write the statement that the equation is continuous on IR – user271716 Sep 18 '15 at 15:34
• What? I don't follow... – Zach466920 Sep 18 '15 at 15:40

The Newton polygon tells us that the dominant binomials are

• $x^5-1102x^4$ for large roots, resulting in a root close to $1102$ and
• $-1102x^4-2015x$ for small roots, resulting in roots close to $0$ and the three third roots of $-\frac{2015}{1102}\approx (-1.22282549628)^3$.

Single real roots stay real under small perturbations, thus giving exactly 3 real roots and a pair of complex conjugate roots. Indeed the numerical approximations confirm this, they are (thanks to http://www.akiti.ca/PolyRootRe.html):

 0
0.6111860961336238   +   1.0593896464200445 i
0.6111860961336238   -   1.0593896464200445 i
-1.2223736979388697
1102.0000015056714