Question

How do you use the method: Completing the Square? Please state the steps.

Answer 1

do you have an example?

Answer 2

Um yes: ax-squared + bx + c = 0. Sorry I cant type the little two on top which is squared.

Answer 3

i can make one up
solve
\[x^2-6x-5=0\]
\[x^2-6x=5\] take half of 6, which is 3, square it and add to both sides to get
\[(x-3)^2=5+9=14\] then take the square root and get
\[x-3=\pm\sqrt{14}\] and finally
\[x=3\pm\sqrt{14}\]

Answer 4

if you do the same prodecure with
\[ax^2+bx+c=0\] you get the quadratic formula
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 5

same as what I do :D

Answer 6

i can write out the steps if you like

Answer 7

\[ax^2+bx+c=0\] subtract c
\[ax^2+bx-c\] divide by a
\[x^2+\frac{b}{a}x=-\frac{c}{a}\] take half of the middle term and square it, add to both sides get
\[(x+\frac{b}{2a})^2=-\frac{c}{a}+\frac{b^2}{4a^2}=\frac{b^2-4ac}{4a^2}\] take the square root get
\[x+\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\pm\frac{\sqrt{b^2-4ac}}{2a}\]

Answer 8

and finally
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 9

Hm... Well, After you get ax-squared + bx = -c, I got confused. I believe my teacher taught me a simpler looking way, but still the same. Right now I'm doing CST Review, so the answer choices are complicated. Is the second step possibly like : x-square + bx = -c/x?

Answer 10

=-c/a

Answer 11

there was a typo in my second step
should have been
\[ax^2+bx=-c\]

Answer 12

Thats the first step in my packet.

Answer 13

??