SOLUTION: Solve algebraically and check your potential solutions: squareroot:x+2-x=0

Algebra ->  Square-cubic-other-roots -> SOLUTION: Solve algebraically and check your potential solutions: squareroot:x+2-x=0      Log On


   



Question 175240: Solve algebraically and check your potential solutions: squareroot:x+2-x=0
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
It really depends on whether you mean sqrt%28x%29%2B2-x=0 or sqrt%28x+%2B+2%29+-+x=0. You could also have meant sqrt%28x+%2B+2+-x%29=0 but that reduces to the absurdity that sqrt%282%29+=0, so I am discounting that possibility.

Case 1:

sqrt%28x%29%2B2-x=0

Let t=sqrt%28x%29, then t+%2B+2+-+t%5E2=0

In standard form: t%5E2+-+t+-+2+=+0.

This factors to: %28t+-+2%29%28t+%2B+1%29=0 so t+=+2 or t+=+-1

But t=sqrt%28x%29 so sqrt%28x%29+=+2 or sqrt%28x%29+=+-1 meaning:

x+=+4 or x+=+1

However, since we converted to a higher degree equation in order to solve this, we may have introduced an extraneous (and therefore incorrect) root. Check:

sqrt%284%29+%2B++2+-+4+=+2+%2B+2+-+4+=+0 Checks.

sqrt%281%29+%2B++2+-+1+=+2+%3C%3E+0 Does not check. Extraneous root.

Therefore the solution set is x+=+4

Case 2:

sqrt%28x+%2B+2%29+-+x=0

Add x to both sides:

sqrt%28x+%2B+2%29+=+x

Square both sides:

x+%2B+2+=+x%5E2

Standard Form:

x%5E2+-+x+-+2+=+0

Factor:

%28x+-+2%29%28x+%2B+1%29=+0

Hence x+=+2+ or x+=+-1

Again, since we squared both sides in the solution process, we have the possibility of an extraneous root.

Check:
sqrt%282+%2B+2%29+-+2=+2+-+2+=+0 Checks

sqrt%28-1+%2B+2%29+-+%28-1%29+=+1+%2B+1+=+2+%3C%3E+0 Does not check. Extraneous root.

Solution set is x+=+2