Solve. √2x+6-√x+4=1

Question

Okay the 2x+6 is all under the radical and the x+4 is also under the radical please help

Answer 1

sqrt(2x + 6) - sqrt(x + 4) = 1
[sqrt(2x + 6) - sqrt(x + 4)]^2 = 1^2
(2x + 6) -2sqrt((2x + 6)(x + 4)) + (x + 4) = 1
3x + 10 - 2sqrt(2x^2 + 14x + 24) = 1
2sqrt(2x^2 + 14x + 24) = 3x + 10 - 1
2sqrt(2x^2 + 14x + 24) = 3x + 9
[2sqrt(2x^2 + 14x + 24]^2 = (3x + 9)^2
4(2x^2 + 14x + 24) = 9x^2 + 54x + 81
8x^2 + 56x + 96 = 9x^2 + 54x + 81
9x^2 + 54x + 81 - 8x^2 - 56x - 96 = 0
x^2 + 46x - 15 = 0
(x^2 + 46x) - 15 = 0
(x + 23)^2 - 529 - 15 = 0
(x + 23)^2 = 544
x + 23 = sqrt(544) or = -sqrt(544)
x = -23 + sqrt(544) or x = -23 - sqrt(544) [rejected]

Notice that we reject x = -23 - sqrt(544), as it will cause sqrt(x + 4) to be come a complex number, i.e. square root of a negative number.