document.write( "Question 1160899: In solving this equation, (x-3)/(x+2)=(x+2)/(x-3) you could cross multiply and get (x-3)^2=(x+2)^2 but you can't simply take the square root of each side because then you get x-3=x+2 and hence 0=5 which is wrong. If you multiply (x-3)(x-3) and (x+2)(x+2) and get x^2-6x+9=x^2+4x+4 then it is solved, x=1/2. My question is, WHY can't you take the square root of both sides, as mentioned earlier? Thank you so much for your help. \n" ); document.write( "
Algebra.Com's Answer #784312 by Boreal(15235)\"\" \"About 
You can put this solution on YOUR website!
sqrt(x-3)^2=+/- (x-3)
\n" ); document.write( "sqrt(x+2)^2=+/- (x+2)\r
\n" ); document.write( "\n" ); document.write( "But the original fractions are not defined when x=-2 or x=3
\n" ); document.write( "So when you cross-multiply, you get (x-3)^2, but it is not defined at x=3.
\n" ); document.write( "One has to go back to the original function and look at the domain. \r
\n" ); document.write( "\n" ); document.write( "Remember in completing the square problems, one has +/- results
\n" ); document.write( "you can get to x=1/2 if (x-3)=-(x+2)=-x-2
\n" ); document.write( "then 2x=1 and x=1/2, but that is allowing, if one will, an x-3 from one side, where 3 is in the domain, and x=-2 from the other side, where -2 is part of the domain, if one will, for the numerator. \r
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