Solving $2^{2x^2}-\left(2^3+2^8\right)\cdot 2^{x^2+2x}+2^{11+4x}=0 $

Question

I have an equation that I have to solve:

$$2^{2x^2}-\left(2^3+2^8\right)\cdot 2^{x^2+2x}+2^{11+4x}=0 $$

Becomes

$$2^{2x^2}-33\cdot 2^{x^2+2x+3}+2^{11+4x}=0$$

And I got completely stuck here. I've made my research and discovered the Kronecker method. As far as I understand, it has to do with matrix? I am not intelligent enough to understand Kronecker method, so is there another solution? Or maybe you'll teach me how to use the method?

Answer 1

Divide by $2^{4x}$, then:

$$2^{2x^2-4x}-\left(2^3+2^8\right)\cdot 2^{x^2-2x}+2^{11+4x}=0 $$

$$2^{2(x^2-2x)}-\left(2^3+2^8\right)\cdot 2^{x^2-2x}+2^{11+4x}=0 $$

and then plug in $u=2^{x^2-2x}$:

$$u^2-(2^3+2^8)u+2^{11}=0.$$

Thereby $u\in\{2^3,2^8\}$ from which the exponent $x^2-2x\in\{3,8\}$. You then have two quadratic equations for $x$ from which to get the roots. Continue accordingly.

Answer 2

try to put $y=2^{x^2}, z=2^{2x}$ then after factorization you'll get $(y-2^8z)(y-2^3z)=0$
that means that $ (y-2^8z)=0$ or $(y-2^3z)=0$ and then substitute $y$ and $z$ you'll find equations of power of two take $log_2$ and then solve all the equations you find.