5

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.

user376343
  • 8,311
user271716
  • 131
  • 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
  • 1
    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

5 Answers5

4

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 $f=0$

chandu1729
  • 3,801
3

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.

Christian Blatter
  • 226,825
3

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 $f(x)$.

NoChance
  • 6,427
1

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

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)

Zach466920
  • 8,341
1

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
Lutz Lehmann
  • 126,666