How do you solve $4^{2 x} + 4^{x} + 12 = 0$ $4^(2x) + 4^x + 12 = 0$ ?

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How do you solve $4^{2 x} + 4^{x} + 12 = 0$ $4^(2x) + 4^x + 12 = 0$ ?

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Answer 1 · 2015-08-05T16:52:39

There is no real solution.

Explanation:

$4^{2 x} + 4^{x} + 12 = 0$ $4^(2x) + 4^x + 12 = 0$

${(4^{x})}^{2} + 4^{x} + 12 = 0$ $(4^x)^2 + 4^x + 12 = 0$

You can do a replacement: $u = 4^{x}$ $u = 4^x$

$u^{2} + u + 12 = 0$ $u^2 + u +12=0$

$u = \frac{- 1 \pm \sqrt{1^{2} - 4 (1) (12)}}{2}$ $u = (-1+-sqrt(1^2-4(1)(12)))/2$

$= \frac{- 1 \pm \sqrt{- 47}}{2}$ $= (-1+-sqrt(-47))/2$

$u$ $u$ is imaginary so we would need $4^{x}$ $4^x$ is imaginary and whie that is certainly possible for imaginary $x$ $x$ , it is not normally handled in a precalculus study of mathematics.

There is no real solution.

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How do you solve $4^{2 x} + 4^{x} + 12 = 0$ $4^(2x) + 4^x + 12 = 0$ ?

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