x²·4^√(2-x) + 4^(2-x) = 4^(√(2-x)+2) + x²·2^(-2x)?

Question

Here's how it looks in proper notation:
https://dl.dropbox.com/u/17850283/yanswers_equation.png

Please show how you got the solutions.

Morewood · Accepted Answer

So it looks like:

x²·4^√(2-x) + 4^(2-x) = 4^(√(2-x)+2) + x²·2^(-2x)

⇒   x²·4^√(2-x) + 4^(2-x) - 4^(√(2-x)+2) - x²·2^(-2x) = 0  {on one side}

⇒   x²·4^√(2-x) + 16[4^(-x)] - 16[4^√(2-x)] - x²·4^(-x) = 0  {exponent laws}

⇒   x²·4^√(2-x) - x²·4^(-x) + 16[4^(-x)] - 16[4^√(2-x)] = 0  {commutativity}

⇒   x²·[4^√(2-x) - 4^(-x)] + 16[4^(-x) - 4^√(2-x)] = 0  {distributive law}

⇒   (x²-16)·[4^√(2-x) - 4^(-x)] = 0  {distribute law again}

⇒   (x-4)·(x+4)·[4^√(2-x) - 4^(-x)] = 0  {difference of squares}

⇒   x=4   or   x=-4   or   √(2-x) = -x {with x≤0 since √(...)≥0}

Square both sides of that last equation to get 0 = x² + x - 2 = (x+2)(x-1) ⇒ x=-2.  (The x=1 solution to that last equation is extraneous.  It does not satisfy the original equation which has only positive square roots.)

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x²·4^√(2-x) + 4^(2-x) = 4^(√(2-x)+2) + x²·2^(-2x)?

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