Equation , powers of two

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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

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

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

Answer 4

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):