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This equation have two real roots, the task is to find them all!

$x^8+x^6+x^4+x^2=340$

Please do not use check and guess or brute force

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1  
So what is your question? –  Matthew Conroy Apr 24 '12 at 21:25
2  
Victor, avoid subjective titles. What might be very difficult for some, might prove easy for others. –  Pedro Tamaroff Apr 24 '12 at 21:26
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Well, you could apply some Galois theory to decompose into irreducible factors, calculate their Galois groups, and check if any of them are soluble. Or just test $x=2$. I'd go for the latter. –  George Lowther Apr 24 '12 at 21:41
    
@GeorgeLowther - Appreciate very much, i really want to shake your hand –  Victor Apr 24 '12 at 21:42
    
Ok, a slightly simpler method would be to substitute $y=x^2$ and apply the standard methods for solving a quartic. But still, I'd check if it has integer solutions first. There's not many possibilities to try. –  George Lowther Apr 24 '12 at 21:43

5 Answers 5

up vote 2 down vote accepted

You can factor the LHS of $x^8+x^6+x^4+x^2=340\Rightarrow x^8+x^6+x^4+x^2-340=0$ to

$$(-2+x) (2+x) (85+21 x^2+5 x^4+x^6)=0$$

Thus the real roots are $\pm 2$.

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7  
And how did you find that factorization, except by already knowing/guessing the real roots? (It is clear that there are only two real roots, because $x^8+x^6+x^4+x^2$ is a sum of convex functions and therefore itself convex). –  Henning Makholm Apr 24 '12 at 21:58

Substitute $y = x^2$, then factor both sides: $$y(y+1)(y^2+1)=(2)(2)(5)(17)$$

If there are to be integer solutions, one of the factors on the left must correspond to 17; it is easy to see that this will be $y^2+1$.

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It is not very difficult. Note that $x=2$ and $x=-2$ are solutions. So you can factor

$$ x^8+x^6+x^4+x^2-340=(x^2-4)Q(x)$$

Make the division and note that $Q$ has positive coefficients, and only even degree terms.

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Is there another way to do this since i hate guess and check all the time –  Victor Apr 24 '12 at 21:30
    
It is not guess and check... Usually you should search for simple solutions $0,\pm 1,\pm 2,...$ but if you do not find small roots it is very unlikely to find some greater roots. Still, if you do find some simple roots, the problem simplifies. –  Beni Bogosel Apr 24 '12 at 21:34
    
Well, i don't know if x is integer(it is not given), maybe it is a integer inside a radical... –  Victor Apr 24 '12 at 21:36
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Seems like its a case of guessing what the questioner deems is an acceptable method and checking if he likes it or not. –  George Lowther Apr 24 '12 at 21:36
    
Another reason why 'guessing' is a good strategy. Generally we cannot solve exactly every polynomial equation of degree greater than 4. So if you are asked to solve the equation you have to guess some solutions or factor the equation such that the factors have degrees smaller than 4. Since few people know by heart the formulas for solutions of the 3rd and 4th degree equations, finally you'd need to reduce the problem to equations of degree at most 2, and eventually equations which have no solutions, like it is in this case. –  Beni Bogosel Apr 24 '12 at 21:46

This is neither the fastest nor the nicest way to solve this problem. In fact, it's the worst way that I can think of to solve this problem by most standards, but it's certainly not guess and check. I started typing this without realizing how long the method would take after the OP asked for a non-guess-and-check solution. By the time I had finished of course others had answered it, but I figured I'd post it anyway because it's mildly amusing.

Note that the equation has only even powers of x. Set $y=x^2$. So we want to find the roots of the polynomial:

$y^4+y^3+y^2+y-340$

This is a quartic polynomial, and the general solution to the quartic equation is well-known. You will get 2 real and 2 nonreal roots. If you work at it you can probably get your roots to look like these:

$y_1=4$

$y_2=\frac{1}{3}(-5-\frac{19\sqrt[3]{4}}{\sqrt[3]{3 \sqrt{19302}-400}}+{\sqrt[3]{2(3 \sqrt{19302}-400)})}$

$y_{3,4}=\frac{1}{3}(-1-\frac{(1\pm\sqrt{3}i)\sqrt[3]{3 \sqrt{19302}-400}}{\sqrt[3]{4}}+\frac{1\mp\sqrt{3}i}{\sqrt[3]{2(3 \sqrt{19302}-400)}})$

Now $y=x^2$ and you want to find the real values of $x$ which solve the equation. Since $x$ was real, this corresponds to $y$ being real and nonnegative. Set $\alpha=\sqrt[3]{3 \sqrt{19302}-400}$. You should check that $\alpha$ is real and positive, this follows from the facts that $134^2=17956<19302$ and $3*134>400$. It's not terribly hard to see that $y_3$ and $y_4$ aren't real. We get that:

$\Im(y_3)=-\frac{1}{\sqrt{3}}(\frac{\alpha}{\sqrt[3]{4}}+\frac{1}{\alpha\sqrt{2}})$,

which is clearly not zero since $\alpha$ is real. To show that $y_2$ is negative, what we need to show is that $\sqrt{2}\alpha < 5+ \frac{19\sqrt[3]{4}}{\alpha}$ which will follow immediately from the facts that $\alpha >0$ (above), $\sqrt{2}<1.5$ (well-known), and $\alpha<3$ (not yet proven), since then $\sqrt{2} \alpha < 1.5*3=4.5<5<5+ \frac{19\sqrt[3]{4}}{\alpha}$. To show that $\alpha<3$ just note that $19302<19600=140^2$ so $\alpha=\sqrt[3]{3 \sqrt{19302}-400}<\sqrt[3]{3*140-400}=\sqrt[3]{20}<\sqrt[3]{27}=3$.

Hence, the only nonnegative real solution for $y$ is $y_1=4$ which correspond to the real roots $x=\pm\sqrt{4}=\pm2$.

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Simply using Rational root theorem yields a root at $+2$. Now we can use Ruffini's rule to find the other one.

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