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
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
Simply using Rational root theorem yields a root at $+2$. Now we can use Ruffini's rule to find the other one.
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 $y^2+1$.
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.
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 $\pm 2$.
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 $x=\pm\sqrt{4}=\pm2$.