1/2 radical 3x+6 = 1 + radical x-6?

Question

(radical is over 3x+ 6)
(radical is over x-6)

how do i solve the equation?

Answer 1

1/2√(3x+6) = 1 + √(x - 6)
√(3x+6) = 2 + 2√(x - 6)

Square both sides:
3x + 6 = (2 + 2√(x - 6))^2

Distribute and solve:
3x + 6 = 4 + 4(x - 6) + 8√(x - 6)
3x + 6 = 4x - 20 + 8√(x - 6)
26 - x = 8√(x - 6)

Square both sides again:
((26 - x))^2 = 64(x - 6)
676 - 52x + x^2 = 64x - 384
1060 - 116x + x^2 = 0

Quadratic formula: x = (-b +/- √(b^2 - 4ac)) / 2a
x = (116 +/- √(13456 - 4240)) / 2
x = (116 +/- 96)/2
x = 10, 106

Check:
1/2√(3x+6) = 1 + √(x - 6)
1/2√(3(10)+6) = 1 + √(10 - 6)
1/2√(36) = 1 + √(4)
1/2 * 6 = 1 + 2
3 = 3

1/2√(3(106) + 6) = 1 + √(106 - 6)
1/2√324 = 1 + √100
1/2 * 18 = 1 + 10
9 = 11 (Thus, ignore this answer).

So, x = 10.