SOLUTION: Solve. x^2+sqrt22x=0

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Question 246147: Solve.
x^2+sqrt22x=0

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2Bsqrt%2822x%29=0
Here's a couple of solutions:
1) A longer, methodical approach.
1a) Eliminate the square root by isolating it and squaring both sides:
x%5E2+=+-sqrt%2822x%29
%28x%5E2%29%5E2+=+%28-sqrt%2822x%29%29%5E2
x%5E4+=+22x
1b) Solve this equation by getting one side equal to zero, factoring and using the Zero Product Property:
x%5E4+-+22x+=+0
x%28x%5E3+-+22%29+=+0
x+=+0 or x%5E3+-+22+=+0
x+=+0 or x+=+root%283%2C+22%29
1c) Whenever you square both sides of an equation, like we did here, you have to check your answers. This is so because squaring both sides of an equation may introduce what are called extraneous solutions. These are solutions which work in the squared equation but not work in the original equation. We are only interested in solutions to the original equation.

x%5E2%2Bsqrt%2822x%29=0
Checking x = 0:
%280%29%5E2%2Bsqrt%2822%280%29%29=0
0%2Bsqrt%280%29=0
0+%2B+0+=+0 Check!

Checking x+=+root%283%2C+22%29:
%28root%283%2C+22%29%29%5E2%2Bsqrt%2822%28root%283%2C+22%29%29%29=0
It will help the check if we use fractional exponents:
%28%2822%29%5E%281%2F3%29%29%5E2+%2B+%2822%2822%5E%281%2F3%29%29%29%5E%281%2F2%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%282%2F3%29+=+0
These are two positive numbers which cannot add up to zero. So x+=+root%283%2C+22%29%29 is NOT a solution.

2) A short, clever solution:
x%5E2 is never negative, It can only be positive or zero.
Square roots are never negative. They can only be positive or zero.
Adding two such numbers and getting zero is only possible if both numbers are zero. (If this is not clear, think about it. If either number is positive then how do we get back to zero when we add the other number?). So the only solution will be x values that make both terms zero:
x%5E2+=+0 and sqrt%2822x%29+=+0
I hope it is obvious that the only x value that will make both if these zero is zero itself. So x=0 is the only solution.