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 $3-sqrt(44)$

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