How do you solve #u^2-4u=2u+35# by completing the square?

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Answer 1 · 2017-02-20T20:58:45

#u=3+sqrt(44)# and #3-sqrt(44)#

First, subtract #2u# on both sides.

#u^2-6u=35#

Now find #(b/(2a))^2# where #a# is the coefficient in front of #u^2# and #b# is the coefficient in front of #u# (so #b=-6# and #a=1# in this case)

#(-6/(2(1)))^2=9#

Now, complete the square by adding both sides by 9.

#u^2-6u+9=44#

Rewrite the left side:

#(u-3)^2=44#

Solve for u. Remember that taking the square root of both sides will give you a positive and negative number.

#(u-3)=sqrt(44)# and #-sqrt(44)#

#u=3+sqrt(44)# and #3-sqrt(44)#