SOLUTION: Solve x^2 + sqrt 22x = 0

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Question 172238This question is from textbook Introductory Algebra
: Solve
x^2 + sqrt 22x = 0
This question is from textbook Introductory Algebra

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
Solve
x%5E2+%2B+sqrt%2822x%29+=+0

Isolate the radical term by subtracting x%5E2 from both sides:

sqrt%2822x%29+=+-x%5E2

Square both sides:

%28sqrt%2822x%29%29%5E2=%28-x%5E2%29%5E2

22x=x%5E4

22x-x%5E4=0

x%2822-x%5E3%29=0



We must always check even-root radical equations,
because there may be extraneous solutions,

Checking x=0

x%5E2+%2B+sqrt%2822x%29+=+0
0%5E2+%2B+sqrt%2822%280%29%29+=+0
0+%2B+sqrt%280%29+=+0
0+%2B+0+=+0
0=0
root(3,22)
Checking x=root%283%2C22%29

x%5E2+%2B+sqrt%2822x%29+=+0
%28root%283%2C22%29%29%5E2+%2B+sqrt%2822root%283%2C22%29%29+=+0
22%5E%282%2F3%29+%2B+sqrt%2822%5E1%2A22%5E%281%2F3%29%29=0
22%5E%282%2F3%29+%2B+sqrt%2822%5E%281%2B1%2F3%29%29=0
22%5E%282%2F3%29+%2B+sqrt%2822%5E%283%2F3%2B1%2F3%29%29=0
22%5E%282%2F3%29+%2B+sqrt%2822%5E%284%2F3%29%29=0
22%5E%282%2F3%29+%2B+%2822%5E%284%2F3%29%29%5E%281%2F2%29=0
22%5E%282%2F3%29+%2B+22%5E%28%284%2F3%29%281%2F2%29%29=0
22%5E%282%2F3%29+%2B+22%5E%282%2F3%29=0
2%2A22%5E%282%2F3%29=0

That is false so x=root%283%2C22%29 is not a
solution.

So x=0 is the only solution.

-------------------

Here is how we could have solved this much easier:

x%5E2+%2B+sqrt%2822x%29+=+0

Isolate the radical term by subtracting x%5E2 from both sides:

sqrt%2822x%29+=+-x%5E2

A square root radical always represents a non-negative number,
so the left side is non-negative.  The right side, however, will
always be negative unless x=0.  So 0 is the only possible
solution, and we check it as above and find it is a solution,
so the only solution is x=0.

Edwin