# How do you solve 2^(x^2+x) - 4^(1+x) = 0?

Aug 9, 2015

Express both terms as powers of $2$ and rearrange to get a quadratic in $x$, giving solutions: $x = - 1$ or $x = 2$

#### Explanation:

Add ${4}^{1 + x}$ to both sides to get:

${2}^{{x}^{2} + x} = {4}^{1 + x} = {\left({2}^{2}\right)}^{1 + x} = {2}^{2 \left(1 + x\right)} = {2}^{2 x + 2}$

So ${x}^{2} + x = 2 x + 2$

Subtract $2 x + 2$ from both sides to get:

$0 = {x}^{2} - x - 2 = \left(x - 2\right) \left(x + 1\right)$

So $x = - 1$ or $x = 2$