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

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Answer 1 · 2017-02-22T16:50:13

#{8}#

You're going to want to get rid of the 1/2 exponent. You can do this by squaring both sides.

#((2x)^(1/2))^2 = (x - 4)^2#

#2x = (x - 4)(x -4)#

#2x = x^2 - 4x - 4x + 16#

#0 = x^2 - 10x + 16#

#0 = (x - 8)(x - 2)#

#x = 8 and 2#

This is a rational equation, so extraneous solutions are always a possibility, if not a likelihood. Always check your solutions.

#(2 * 2)^(1/2) =^? 2 - 4#

#sqrt(2) != -2 color(red)(xx)#

#(2 * 8)^(1/2) =^? 8 - 4#

#sqrt(16) = 4#

The only valid solution is #x = 8#.

Hopefully this helps!