3 – sqrt[x] = sqrt[2x-7] square both sides
(3-sqrt[x])(3-sqrt[x]) = ([2x-7]^0.5)^2 multiply out on left, mulitply exponents on the right
3*3 – 3*sqrt[x] – 3*sqrt[x] + sqrt[x]*sqrt[x] = [2x-7]^1 simplify
9 – 6*sqrt[x] + x = 2x – 7 subtract 9 and x from both sides
-6*sqrt[x] = x – 16 square both sides
36x = x^2 – 32x + 256 subtract 36x from both sides
x^2 – 68x + 256 use quadratic
get x = 4 or 64
sqrt[x] = y – 5 square both sides
x = y^2 – 10y + 25 then
x + y = 11 subtract y from both sides
x = 11 – y substitute “x = 11 – y” in x = y^2 – 10y + 25
11 – y = y^2 – 10y + 25 subtract 11 and add y to both sides
0 = y^2 – 9y + 14 use quadratic to solve and get
y = 2 or 7 go back to “x + y = 11”
x = 9 or 4
No, 64 is still correct because the square root of 64 can be -8 or +8. If you take the square root of 64 to be -8 then you have 3 – -8 = 11 = sqrt(121)