sqrt(2w + 8) = w where w is a real number?

√(2w + 8) = w
==> [√(2w + 8)]^2 = w^2
==> 2w + 8 = w^2
==> w^2 - 2w - 8 = 0
==> (w - 4)(w + 2) = 0
==> w = 4, w = -2

Check:

-> w = 4 <-
√(2w + 8) = w
==> √[2(4) + 8] = 4
==> √(8 + 8) = 4
==> √16 = 4
==> 4 = 4

-> w = -2 <-
==> √(2w + 8) = -2
==> √[2(-2) + 8] = -2
==> √(-4 + 8) = -2
==> √4 = -2
==> 2 ≠ -2

w = -2 is extraneous. Therefore, the only solution is w = 4.

Hope this helps!

√(2w + 8) = w
2w + 8 = w^2
w^2 - 2w - 8 = 0
w^2 + 2w - 4w - 8 = 0
(w^2 + 2w) - (4w + 8) = 0
w(w + 2) - 4(w + 2) = 0
(w + 2)(w - 4) = 0

w + 2 = 0
w = -2 {√[2(-2) + 8] = √(-4 + 8) = √4 ONLY equal to 2}

w - 4 = 0
w = 4

∴ w ONLY equal to 4.

2w+8=w^2
w^2-2w-8=0
w=[1+&-(1+8)^1/2]=4 & -2,both MUST be checked
\/''16'''=4,w=4 is one answer
\/'''4'''=-2 w=-2 is NOT an answer.

w= 4

Sqrt(2(4)+8) = 4

Sqrt (16) = (4)