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.

$$4\left(a^2+\frac1{a^2}\right)-23\left(a+\frac1a\right)+40=0$$
$$\implies 4\left[\left(a+\frac1a\right)^2-2\right]-23\left(a+\frac1a\right)+40=0$$
$\displaystyle a+\frac1a=b\implies 4(b^2-2)-23b+40=0\iff 4b^2-23b+32=0$
