# Equation , powers of two

I want to find the sum of the roots of the equation $$4(4^x + 4^{-x}) - 23(2^x + 2^{-x}) + 40 = 0$$ in real numbers. I tried the substitution $2^x = t$ but then it turns into a quartic equation which I couldn't solve. I think its roots sum to zero so I want to prove it without actually finding the roots.

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$$(2^x+2^{-x})^2=4^x+2+4^{-x}$$ may help here. –  Oleg567 Jun 24 '14 at 10:43

If $x$ is a root of this equation, then $-x$ is too. Thus, the sum of the roots is $0$.

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Your argument assumes that there is at least one solution. –  gammatester Jun 24 '14 at 10:47
@gammatester: Nope. If there are no roots, the sum of no roots is still $0$. (While we're at it, if $0$ is a root, $-0$ isn't a separate root, but the contribution to the sum is still $0$.) –  user2357112 Jun 24 '14 at 10:48

Let $2^x+2^{-x}=t$. Then $4(t^2-2)-23t+40=0 \implies 4t^2-23t+32=0$

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$$4\left(a^2+\frac1{a^2}\right)-23\left(a+\frac1a\right)+40=0$$

$$\implies 4\left[\left(a+\frac1a\right)^2-2\right]-23\left(a+\frac1a\right)+40=0$$

$\displaystyle a+\frac1a=b\implies 4(b^2-2)-23b+40=0\iff 4b^2-23b+32=0$

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First of all, if $\exists{r}\in\mathbb{R}$ such that $x=r$ is a solution, then $x=-r$ is also a solution.

So the sum of the roots of the equation (if there are any) is obviously $0$.

Second, here is a full analysis of your equation (from this link):

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