How do you solve #\sqrt { 4u + 5} = u#?

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Answer 1 · 2016-12-14T00:54:39

#u=5#

Given:

#sqrt(4u+5) = u#

First square both sides (noting that this may introduce spurious solutions) to get:

#4u+5 = u^2#

Subtract #4u+5# from both sides to get:

#0 = u^2-4u-5 = (u-5)(u+1)#

Hence #u=5# or #u=-1#

The value #u = -1# is not a solution of the original equation, since:

#sqrt(4(-1)+5) = sqrt(1) = 1 != -1#

The value #u=5# is a solution of the original equation:

#sqrt(4(5)+5) = sqrt(25) = 5#