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I got this question on my homework and I cannot for the life of me figure out how to solve for $0$.

$$x^5+2x-10=0$$

I have tried this every which way and this is my last resort. Thanks in advanced.

Amzoti
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CoffeePoweredComputers
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  • I would like the real solution please and a brief expelenation. :) – CoffeePoweredComputers Mar 08 '15 at 05:53
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    No im just a pre calc student who is obsessed with doing everything manually :) – CoffeePoweredComputers Mar 08 '15 at 05:56
  • I know this is alot to ask but would you mind showing me how to do it with this problem? – CoffeePoweredComputers Mar 08 '15 at 05:58
  • To be more specific: Let f(x) = x^5+2x-10 for short. Then f(2) > 0 and f(1) < 0, so the graph must cross the x-axis (so f must equal 0) somewhere in between. Now get out a calculator. What is f(1.5)? Whether it's positive or negative, you have now "trapped" the solution in an interval of length 0.5. Keep going, and you can get it as accurate as you like provided you have the patience! – John Brevik Mar 08 '15 at 06:02
  • Thanks everyone! Ill give it a shot. :) – CoffeePoweredComputers Mar 08 '15 at 06:04
  • See Abel-Ruffini theorem. – Lucian Mar 08 '15 at 06:19
  • @Amzoti: I don't think this equation has any imaginary roots. Indeed, the real part is $-10$ for any imaginary $x$. – tomasz Mar 08 '15 at 11:19
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    @tomasz: I'm sure Amzoti meant "complex" instead of "(purely) imaginary". It's a common enough (abuse of) terminology that I didn't even notice it before you drew attention to it. Indeed, the use of "imaginary number" to mean "complex number that is not purely real" goes back to Descartes himself, and is the reason why the term "purely imaginary number" exists. – Ilmari Karonen Mar 08 '15 at 13:09
  • @IlmariKaronen: Sure. Still, I think it's better to say "nonreal" (when there is no risk of confusion in a different direction: there are nonreal and noncomplex quaternions, for example; of course, quaternions with zero real part are also called imaginary, so there's no way to avoid verbosity in cases like that). – tomasz Mar 08 '15 at 14:22
  • @Davsmith4 That attitude of doing everything manually is (mostly) good. At some point, it might get boring, but it's important to understand how things work, which is what doing them manually does. Besides, if you do hard problems manually, you might invent shortcuts that you can use (my favorite thing; I use these all the time). – Justin Mar 08 '15 at 19:32
  • Does your homework actually say "solve for 0"? Because "solving for ____" means expressing an equation in terms of ____. This may just be a trick question, where your teacher wants you to write "$0 = x^5 + 2x - 10$" or perhaps even "$0 = 0$". – ruakh Mar 08 '15 at 19:45
  • No I was trying to find the roots of a this x^7+2x^3-10x^2 to graph it and after factoring out x^2 and solving that for 0 I was left with my question above. – CoffeePoweredComputers Mar 08 '15 at 19:58

4 Answers4

14

If $$f(x)=x^5+2x-10$$ $$f'(x)=5x^4+2 >0$$ so there is only one real root since the function varies from $-\infty$ to $+\infty$ because of $x^5$.

By inspection, you can notice that $f(1)=-7$ and $f(2)=+26$ so the solution is between $1$ and $2$.

To find the solution, Newton method is quite simple : starting from a "reasonable" guess $x_0$, the method will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ In the present case, this will the simply write $$x_{n+1}=\frac{4 x_n^5+10}{5 x_n^4+2}$$ Let us be lazy and start at the middle of the interval then using $x_0=1.5$. This will generate the following iterates : $1.47826$, $1.47765$ which is the solution for six significant digits.

Edit

You could get a quite accurate approximation using Taylor series built at $x=\frac 32$. Limited to second order, this gives $$x^5+2x-10=\frac{19}{32}+\frac{437}{16} \left(x-\frac{3}{2}\right)+\frac{135}{4} \left(x-\frac{3}{2}\right)^2+O\left(\left(x-\frac{3}{2}\right)^3\right)$$ and the solution of the quadratic is $$\frac{1183+\sqrt{170449}}{1080}\approx 1.477643$$ while the exact solution is $\approx 1.477653$.

Claude Leibovici
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  • I got pretty much all of this except how did you get the top part of the newton method? – CoffeePoweredComputers Mar 08 '15 at 19:15
  • @Davsmith4 $\displaystyle x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{x_n^5+2x_n-10}{5x_n^4+2}=\frac{x_n (5x_n^4+2)}{5x_n^4+2}-\frac{x_n^5+2x_n-10}{5x_n^4+2}=\frac{5x_n^5+2x_n-(x_n^5+2x_n-10)}{5x_n^4+2}=\frac{4x_n^5+10}{5x_n^4+2}$ – Justin Mar 08 '15 at 19:27
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Your polynomial has a unique real solution which you can estimate numerically to be around $1.478$. PARI computes the Galois group to be $S_5$, which means that you can't express this solution with radicals (roots).

Yuval Filmus
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    I like that you provide that the Galois group of this polynomial is $S_5$, because the other answers don't note that there is no solution expressible by radicals. – Justin Mar 08 '15 at 19:30
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Differentiate and you will see that the function $f(x) = x^5+2x-10$ is strictly increasing.

It is easy to see that $\lim_{x \to -\infty} f(x) = -\infty$ and $\lim_{x \to \infty} f(x) = \infty$, so there is exactly one root.

Since $f(0) <0$ and $f(2) >0$ we see that the root lies in $(0,2)$.

You can use bisection to find a numerical approximation to the root.

Using 10 iterations of the bisection method starting with the interval $[0,2]$ gives the bracket $(1.4765625, 1.478515625)$.

copper.hat
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Just for completeness. The real solution of this equation can be expressed in terms of a generalized hypergeometric function

$$ 5 \, {}_4F_3\!\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};-\frac{1953125}{512}\right).$$

yarchik
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