x = 4 is the answer, if you take the positive square roots. There may be other answers if you take the negative square root on one or both sides.
I got that by trial and error, not by using algebra. The algebra got too messy.
solve:
the square root of (x) +2= the square root of (5-x) +3
3 answers
Whatever roots exist must be between x = 0 and 5, otherwise you get imaginary numbers. As I said in a previous post, x=4 is the answer is you take postive squate roots on both sides. x=1 is also an answer, if you take the positive root of 1 on the left and the negative root of 4 on the right.
Here is the algebraic proof:
sqrt(x) + 2 = sqrt(5-x) + 3
sqrt(x) -1 = sqrt(5-x)
2x = 2 sqrt(x) +4
x = sqrt(x) + 2
Let sqrt(x) = y, and you have a quadratic equation:
y^2 -y +2 = (y-2)(y+1) = 0
y = 2 or -1
x = y^2 = 4 or 1
sqrt(x) + 2 = sqrt(5-x) + 3
sqrt(x) -1 = sqrt(5-x)
2x = 2 sqrt(x) +4
x = sqrt(x) + 2
Let sqrt(x) = y, and you have a quadratic equation:
y^2 -y +2 = (y-2)(y+1) = 0
y = 2 or -1
x = y^2 = 4 or 1
4 answers