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.
<pre>{{{matrix(1,3, (x - 3)/(x + 2), "=", (x + 2)/(x - 3))}}}
{{{matrix(1,3, (x - 3)^2, "=", (x - 2)^2)}}} ----- Cross-multiplying
A lot of people <b><u>FORGET</b></u>, I believe, that when taking the square root of an expression, it's IMPERATIVE to indicate that the resulting expression can either be - (negative), or + (positive). 
Taking the square root of both sides, we get: {{{matrix(1,3, x - 3, "=", ""+- (x + 2))}}}
{{{matrix(1,3, matrix(2,3, x - 3, "=", x + 2,
- 3, "=", 2), "<====", highlight(FALSE))}}}      {{{matrix(1,3, matrix(5,3, x - 3, "=", - (x + 2),
x - 3, "=", - x - 2,
x + x, "=", - 2 + 3,
2x, "=", 1,
x, "=", 1/2), "<====", highlight(TRUE))}}}
<b><u>OR</b></u>
Taking the square root of both sides, we get: {{{matrix(1,3, x + 2, "=", ""+- 
(x - 3))}}}
{{{matrix(1,3, matrix(2,3, x + 2, "=", x - 3,
2, "=", - 3), "<====", highlight(FALSE))}}}      {{{matrix(1,3, matrix(5,3, x + 2, "=", - (x - 3),
x + 2, "=", - x + 3,
x + x, "=", 3 - 2,
2x, "=", 1,
x, "=", 1/2), "<====", highlight(TRUE))}}}
Another VERY IMPORTANT fact is that {{{matrix(1,3, x <> - 2, and, x <> 3)}}}