How do you solve #u=sqrt(-u+12)#?

Question

1410 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-29T07:02:39

#u = - 4 or u = 3#

Follow these simple steps

#u = sqrt (-u + 12)#

Square both sides

#u^2 = sqrt (-u + 12)^2#

#u^2 = -u + 12#

Collect like terms

#u^2 + u - 12 = 0 -> Quadratic#

Solving the Equation we have #+4 , -3# as the quadratic factors

#u^2 + 4u - 3u - 12 = 0#

#(u^2 + 4u) (- 3u - 12) = 0#

#u(u + 4) -3 (u + 4) = 0#

#(u + 4) (u -3) = 0#

#(u + 4) = 0 or (u -3) = 0#

#u = - 4 or u = 3#