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

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Answer 1 · 2015-08-09T22:21:42

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

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

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

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

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

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

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

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