SOLUTION: Hi, the equation is: X - 4*(square root X) = 32. I thought about squaring both sides to get a quadratic, but that would still leave a (square root of x) in the equation. Thanks

Algebra ->  Radicals -> SOLUTION: Hi, the equation is: X - 4*(square root X) = 32. I thought about squaring both sides to get a quadratic, but that would still leave a (square root of x) in the equation. Thanks       Log On


   

Question 829204: Hi, the equation is: X - 4*(square root X) = 32. I thought about squaring both sides to get a quadratic, but that would still leave a (square root of x) in the equation. Thanks for any help you can provide.
Found 2 solutions by edjones, KMST:
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
x-4sqrt(x)=32
x-4sqrt(x)-32=0
(sqrt(x)-8)(sqrt(x)+4)=0
sqrt(x)=8
x=64
.
Ed

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
If there is only one square root, you want to isolate it on one side of the equation before squaring.
x-4sqrt%28x%29=32
-4sqrt%28x%29=32-x or x-32=4sqrt%28x%29
Also, if sqrt%28x%29 has to be a real number, it must be that x%3E=0 .
Now we can square both sides, knowing that we may be introducing some exraneous solutions.
%28x-32%29%5E2=%284sqrt%28x%29%29%5E2
x%5E2-64x%2B1024=4%5E2%2A%28sqrt%28x%29%29%5E2
x%5E2-64x%2B1024=16x
x%5E2-64x%2B1024-16x=0
x%5E2-80x%2B1024=0
Factoring, since x%5E2-80x%2B1024=%28x-16%29%28x-64%29 , we transform the equation into
%28x-16%29%28x-64%29=0 , whose solutions are
x=16 and x=64
Is any one of those an extraneous solution?
For x=16,
x-4sqrt%28x%29=16-4sqrt%2816%29=16-4%2A4=16-16=0 so x=16 is not a solution of the original equation. It's an extraneous solution, introduced when we squared both sides.
For x=64,
x-4sqrt%28x%29=64-4sqrt%2864%29=64-4%2A8=64-32=32 So highlight%28x=64%29 is a solution to the original equation.