Question

solve for x: 7-(x^2-3x)=sqrt[(x^2-3x)+5)]
I came up with only one solution, but I think there should be four. Only one worked.

Answer 1

\[7-(x^{2}-3)=\sqrt{(x^{2}-3)+5}\]

Answer 2

Let x^2-3 be X so

7 -X = sqrt(X +5) and square to get a quadratic

44 -15X + X^2 = 0 which has roots of 11 or 4.

X cannot be 11 because 7-X is negative and we are looking for the positive square root.

4 works so now substitute X = 4

X = 4 = x^2 -3 or x^2 = 7 so x = plus minus sqrt 7 and these both work because they are squared in the expression.