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????
$\displaystyle x^2+ax$ to complete the square we take half the coeffient on the x term.
So in this case $\displaystyle \frac{a}{2}$ and then we square it
$\displaystyle \left( \frac{a}{2} \right)^2=\frac{a^2}{4}$
Then we add this to get..
$\displaystyle x^2+ax+\frac{a^2}{4}=(x+\frac{a}{2})^2$
Now to apply this to your problem.
$\displaystyle x^2-4x=17$ so we take half the x term and square it $\displaystyle \left( \frac{-4}{2} \right)^2=4$
So now we add 4 to both sides of the equation
$\displaystyle x^2-4x +4=17+4 \iff (x-2)^2=21$ now take the square root of both sides
$\displaystyle x-2=\pm \sqrt{21} \iff x=2 \pm \sqrt{21}$
Good luck
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
so adding 4 we have,
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.