# Thread: Solving by completing the square.

1. ## Solving by completing the square.

Hey I need help on figuring out this problem...and I'm in middle school doing algebra made for people in high school!!! Ok here is the problem:
x²-4x=17
I know to you that probably looks easy, but what do you do????

2. ## to complete the square

$x^2+ax$ to complete the square we take half the coeffient on the x term.

So in this case $\frac{a}{2}$ and then we square it

$\left( \frac{a}{2} \right)^2=\frac{a^2}{4}$

Then we add this to get..

$x^2+ax+\frac{a^2}{4}=(x+\frac{a}{2})^2$

Now to apply this to your problem.

$x^2-4x=17$ so we take half the x term and square it $\left( \frac{-4}{2} \right)^2=4$

So now we add 4 to both sides of the equation

$x^2-4x +4=17+4 \iff (x-2)^2=21$ now take the square root of both sides

$x-2=\pm \sqrt{21} \iff x=2 \pm \sqrt{21}$

Good luck

3. x²-4x=17

Now , to factorize by completing square, we must make the LHS in the form
( a - b )^2 that is a^2 -2ab + b^2

It is found that if we add 4 to the equation then the LHS does become of the above form

x^2 - 4x + 4 = 17 + 4

therefore

(x)^2 - 2(x)(2) + (2)^2 = 21

so, factorizing we get,

(x-2)^2 = 21, transposing 21 we get

[(x-2)]^2 - 21 = 0

This looks similar, infact it imitates the identity , a^2 - b^2 = (a-b)(a+b)

so further factorizing it we get,

[x-2+ ROOT(21) ] [x-2- ROOT(21)]=0

SOLUTIONS, either x = 2- ROOT(21) OR x = 2 + ROOT(21)

hence found.