Question

how do I complete the square?

y= x^2 - 6x + 7

Answer 1

$$(x-3)^2-2=x^2-6x+7$$

Answer 2

how did you get that though?

Answer 3

$$(x-a)^2=x^2-2ax+a^2$$

Answer 4

Subtract/add to the left hand side until the constant on the right is what you want

Answer 5

\[x^2 - 6x + 7 =x^2 - 6x +9-9 +7=x^2 - 6x +3^2-2=(x-3)^2-(\sqrt{2})^2\]

Answer 6

hmm

Answer 7

do you understand?

Answer 8

not really D:

Answer 9

http://www.purplemath.com/modules/sqrquad.htm

look at the example in the box

Answer 10

Dear Laurenn_215, in order to complete the square of the given function check the coeficient of x i.e.(- 6x) it is -6
next find (-6)/2=-3
find the square of (-3) i.e. (-3)^2=9
Now add and subtract 9 in the given function so it will be
\[x^2 - 6x + 7 = x^2 - 6x + 9-9+7 = x^2 - 2\times x \times 3 + 9-2\]
\[= x^2 - 2\times x \times 3 + 3^2-2\]
Now this part i.e. \[x^2 - 2\times x \times 3 + 3^2\] is written in the form of \[a^2 - 2\times a \times b + b^2 = (a-b)^2\]
so x^2 - 2times x times 3 + 3^2-2 can be expressed as \[= (x-3)^2-2\]

Answer 11

thanks everyone