SOLUTION: how do you solve {{{ sqrt(2x)= x-4 }}}

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Question 395017: how do you solve +sqrt%282x%29=+x-4+
Found 2 solutions by ewatrrr, jsmallt9:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi, Note correction of middle term of trinomial**

solving

+sqrt%282x%29=+x-4+   |squaring both sides

  2x = (x-4)^2

  2x = x^2 - 8x + 16

  x^2 -10x + 16 = 0    |**

factoring

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

(x-8)= 0  x = 8

(x-2) = 0 x = 2  Extraneous solution

 x = 8

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Another tutor's solution is mostly correct. But there is a mistake.
+sqrt%282x%29=+x-4+
Square both sides:
+%28sqrt%282x%29%29%5E2=+%28x-4%29%5E2+
2x+=+x%5E2+-8x+%2B+16
(Subtracting 2x from each side (this is where the mistake was made):
0+=+x%5E2+-10x+%2B+16
Factor:
(x-2)(x-8) = 0
From the Zero Product Property we know that this (or any) product can be zero only if one (or more) of the factors is zero. So:
x-2 = 0 or x-8 = 0
Solving these we get:
x = 2 or x = 8

Whenever you square both sides of an equation extraneous solutions may be introduced. Extraneous solutions are solutions that fit the squared equation but do not fit the original equation. These extraneous solutions can occur even if no mistakes have been made! For this reason you must check your answers any time you square both sides of an equation.

When checking, always us the original equation:
+sqrt%282x%29=+x-4+
Checking x = 2:
+sqrt%282%282%29%29=+%282%29-4+
+sqrt%284%29=+-2+
+2+=+-2+ Check failed! This is an extraneous solution and must be rejected.

Checking x = 8:
+sqrt%282%288%29%29=+%288%29-4+
+sqrt%2816%29=+4+
+4+=+4+ Check!

So the only solution to your equation is x = 8.