Solving a quintic function for zero

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.

• I would like the real solution please and a brief expelenation. :) – CoffeePoweredComputers Mar 8 '15 at 5:53
• No im just a pre calc student who is obsessed with doing everything manually :) – CoffeePoweredComputers Mar 8 '15 at 5:56
• I know this is alot to ask but would you mind showing me how to do it with this problem? – CoffeePoweredComputers Mar 8 '15 at 5:58
• 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 8 '15 at 6:02
• @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 8 '15 at 13:09

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 $\approx 1.477653$.

• I got pretty much all of this except how did you get the top part of the newton method? – CoffeePoweredComputers Mar 8 '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 8 '15 at 19:27

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).

• 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 8 '15 at 19:30

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 $(1.4765625, 1.478515625)$.