Question

Hello, I need help intergrating this problem.. I know it involves completing the square but i can't seem how t start. INT dx/8x^2-8x+11

Answer 1

\[8x^2-8x+11 =8(x^2-x)+11=8(x^2-1x)+11\\=8(x^2-1x+(\frac{1}{2})^2)+11-8(\frac{1}{2})^2=8(x-\frac{1}{2})^2+11-2\]

Answer 2

Did you do defraction composition in this problem instead?

Answer 3

You just look at the ax^2+bx part...
\[ax^2+bx+c \\= (ax^2+bx)+c \\ = a(x^2+\frac{b}{a}x)+c \\ =a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2 )+c-a (\frac{b}{2a})^2 \text{ completing the \square part } \\ =a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a} \\ a(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a} \\ =[\sqrt{a}(x+\frac{b}{2a})]^2+\frac{4ac-b^2}{4a}\]

Answer 4

Never heard of the term defraction composition
But I definitely completed the square.

Answer 5

the above was assuming a is positive there

Answer 6

Yes A is positive, I see what you did there just trying to comprehend it now for my test :P

Answer 7

Well I was talking about my general quadratic for which I was completing square for

Answer 8

since I shove the sqrt(a) inside that square think
sqrt(a) exists over the reals when a>0

Answer 9

or equal to 0
but it wouldn't be a quadratic if a=0