SOLUTION: 1. Solve sqrt2x + x = 4 for x. Answers: A. x = 2 B. The two solutions are for complex numbers. C. x = 2 or x = 8 ( I picked C and it was wrong) D. x = 4 or x = -4 Thank Yo

Algebra ->  Radicals -> SOLUTION: 1. Solve sqrt2x + x = 4 for x. Answers: A. x = 2 B. The two solutions are for complex numbers. C. x = 2 or x = 8 ( I picked C and it was wrong) D. x = 4 or x = -4 Thank Yo      Log On


   

Question 1019621: 1. Solve sqrt2x + x = 4 for x.
Answers: A. x = 2 B. The two solutions are for complex numbers.
C. x = 2 or x = 8 ( I picked C and it was wrong) D. x = 4 or x = -4
Thank You.

Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39615) About Me  (Show Source):
You can put this solution on YOUR website!
Your text is not well written so one could guess you have sqrt(2x)+x=4, which would best appear sqrt%282x%29%2Bx=4.

Isolate the radical.
sqrt%282x%29=4-x

Square the equated members.
2x=%284-x%29%5E2
2x=16-8x%2Bx%5E2
Simplify and solve.
x%5E2-8x-2x%2B16=0
x%5E2-10x%2B16=0
%28x-2%29%28x-8%29=0
and if you check you find both x=2 and x=8 will work.
x=2 OR x=8.

Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!

1. Solve sqrt2x + x = 4 for x.
Answers: A. x = 2 B. The two solutions are for complex numbers.
C. x = 2 or x = 8 ( I picked C and it was wrong) D. x = 4 or x = -4
Thank You.
This appears to be sqrt%282x%29+%2B+x+=+4. If so, the answer CANNOT be C, but: highlight%28highlight_green%28highlight%28matrix%281%2C2%2CCHOICE%2C+A%29%29%29%29, instead.

Why don't you use parentheses, or provide info to make this problem less ambiguous? Is that so difficult?