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 #