I want to rewrite $2x^2-12x+11$ in square form. According to my book, the answer is $2(x-3)^2-7$. However, I get $2(x-3)^2+2$ for some reason.
My Steps:
$2x^2-12x+11$
$2x^2-12x=-11$
$2(x^2-6x)=-11$
$2(x^2-6x+9)=-11+9$
$2(x - 3)^2=-2$
$2(x-3)^2+2$
What am I doing wrong?
$\endgroup$ 42 Answers
$\begingroup$As noted in the comments, in the fourth line, you have to add 18 to the right member and not 9.
$2x^2-12x+11=0$
$2x^2-12x=-11$
$2(x^2-6x)=-11$
$2(x^2-6x+9)=-11+18$
$2(x - 3)^2=7$
$2(x-3)^2-7=0$
$\endgroup$ 0 $\begingroup$An alternative way to see how this all works is to note that completing the square doesn't change the value of the expression. What you are essentially doing is adding a cleverly-written $0$ to it so that it becomes easier to work with.
Observe, $$\begin{align}2x^2-12x+11 & = 2x^2 - 12x \qquad\qquad\qquad\qquad\qquad\quad +11 \tag{Leave space}\\ &= 2[x^2 - 6x]\qquad\qquad\qquad\qquad\qquad\;\; +11 \tag{Factor}\\&= 2[x^2 - 6x +\underbrace{\left({-6\over2}\right)^2 - \left({-6\over2}\right)^2}_\text{cleverly-written zero}] + 11 \\ &= 2[\underbrace{x^2 - 6x + 9}_{(x-3)^2} - 9] + 11\\ &= 2[(x-3)^2 - 9] + 11\\ &= 2(x-3)^2 - 18 + 11\\ &= 2(x-3)^2 - 7.\end{align}$$
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