How do you solve $sqrt [w+1]-5=2w$ and find any extraneous solutions?

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Answer 1 · 2016-07-05T12:32:34

$w = -(19)/8 +- (sqrt(23))/8 i$

Add 5 to both sides, giving:

$sqrt(w+1) = 2w + 5$

Square both sides

$w+1 = (2w+5)^2 = 4w^2 + 20w +25$

Collect like terms and solve the quadratic

$4w^2 + 19w + 24 = 0$

$w = (-19+-sqrt(19^2 - 4(4)(24)))/(2(4)) = (-19+-sqrt(-23))/(8)$

$sqrt(-23) = sqrt(23i^2) = sqrt(23)i$

$implies w = -(19)/8 +- (sqrt(23))/8 i$

How do you solve sqrt [w+1]-5=2wsqrt [w+1]-5=2w and find any extraneous solutions?