SOLUTION: What value of x will satisfy this equation {{{sqrt(2x)+4=x}}}

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Question 1042927: What value of x will satisfy this equation sqrt%282x%29%2B4=x
Found 2 solutions by ikleyn, jim_thompson5910:
Answer by ikleyn(52794) About Me  (Show Source):
You can put this solution on YOUR website!
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What value of x will satisfy this equation sqrt%282x%29%2B4=x
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Answer. x = 8.

Solution 

sqrt%282x%29%2B4 = x  --->  sqrt%282x%29 = x-4  --->  square both sides --->

2x = %28x-4%29%5E2  --->  2x = x%5E2+-+8x+%2B+16  --->  x%5E2+-+10x+%2B+16 = 0  --->  factor left side --->

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

The roots are x = 2 and x = 8.

The root x = 8 satisfies the original equation.

The root x = 2 does not satisfy the original equation. It is an extraneous solution.


Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%282x%29%2B4=x


sqrt%282x%29%2B4-4=x-4 Subtract 4 from both sides


sqrt%282x%29=x-4


%28sqrt%282x%29%29%5E2=%28x-4%29%5E2 Square both sides to undo the square root


2x=x%5E2-8x%2B16


2x-2x=x%5E2-8x%2B16-2x


0=x%5E2-10x%2B16


x%5E2-10x%2B16=0


%28x-8%29%28x-2%29=0


x-8=0 or x-2=0


x=8 or x=2


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The two possible solutions are x=8 or x=2


Let's check x=8


sqrt%282x%29%2B4=x


sqrt%282%2A8%29%2B4=8 Replace every x with 8


sqrt%2816%29%2B4=8


4%2B4=8


8=8 Possible solution is confirmed. So x=8 is a true solution


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Let's check x=2


sqrt%282x%29%2B4=x


sqrt%282%2A2%29%2B4=2 Replace every x with 2


sqrt%284%29%2B4=2


2%2B4=2


6=2 This is false, so x+=+2 is extraneous. It is NOT a true solution.


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Final Answer:


x=8 is the only true solution to the original equation.