SOLUTION: I need help please. Solve squrrt2x + x = 4 for x. I got lost, so my answer isn't an option. Not sure where I went wrong. sqrrt2x + x = 4 sqrrt2x -x = 4

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Question 768973: I need help please.
Solve squrrt2x + x = 4 for x.
I got lost, so my answer isn't an option. Not sure where I went wrong.
sqrrt2x + x = 4
sqrrt2x -x = 4 - x
sqrrt2x = 4-x
(squrrt2x)^2 = (4-x)^2
2x = 16-4x-4x+x
2x = 16-8x+x
2x = 16-7x
16-9x
-16/9 = x?
Answer Options:
A. x = 2
B. The two solutions are complex numbers.
C. x = 4 or x = -4
D. x = 2 x = 8
Found 3 solutions by oscargut, lwsshak3, solver91311:
Answer by oscargut(2103)   (Show Source):
You can put this solution on YOUR website!

Solve squrrt2x + x = 4 for x.
I got lost, so my answer isn't an option. Not sure where I went wrong.
sqrrt2x + x = 4
sqrrt2x -x = 4 - x
sqrrt2x = 4-x
(squrrt2x)^2 = (4-x)^2
2x = 16-4x-4x+x (Here is the error)

Answer
A. x = 2

You can ask more at: mthman@gmail.com
Thanks

Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website!
Solve sqrt(2x) + x = 4 for x.

square both sides
2x=16-8x+x^2
x^2-10x+16=0
(x-8)(x-2)=0
x=8
or
x=2

Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!

sqrt(2x) + x = 4
sqrt(2x) + x - x = 4 - x
sqrt(2x) = 4 - x
(sqrt(2x))^2 = (4 - x)^2
2x = 16-4x-4x+x Your problem is in this step.
2x = 16-8x+x
2x = 16-7x
16-9x
-16/9 = x?

Should be:

Collect like terms:

From here, factor and solve. Then check both roots and discard any that do not create a true statement when substituted back into the original equation, since it is possible to introduce extraneous roots when you square both sides of an equation.

John

Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it

SOLUTION: I need help please. Solve squrrt2x + x = 4 for x. I got lost, so my answer isn't an option. Not sure where I went wrong. sqrrt2x + x = 4 sqrrt2x -x = 4 - x sqrrt2x =