Question

Can someone describe me the process of completing the square?

Answer 1

is this to solve an equation or just to complete it?

Answer 2

both the completing part converts standard to vertex right

Answer 3

Here's an excellent tutorial http://www.purplemath.com/modules/sqrquad.htm

Answer 4

??

Answer 5

ax^2+bx+c
half the bx term and square it

Answer 6

to make something vertex you complete the square right

Answer 7

ah yes. but there is an easy trick so you don't have to complete the square

Answer 8

which sometimes is a pain, especially if the leading coefficient is not 1

Answer 9

what is the trick

Answer 10

suppose i want to turn
\[y=ax^2+bx+c \] into
\[a(x-h)^2+k\] so vertex will be (h,k)

Answer 11

for example, suppose i want to turn
\[y=2x^2-6x+9\] into
\[y =2(x-k)^2+h\]

Answer 12

yea

Answer 13

just compute
\[-\frac{b}{2a}\] which in my example is
\[-\frac{-6}{2\times 2}=\frac{3}{2}\]

Answer 14

oh i forgot about the vertex formula thanks

Answer 15

then i write
\[y=2(x-3)^2+h\] and to find h i substitute
\[\frac{3}{2}\] for x to get the answer

Answer 16

then i write
\[y=2(x-3)^2+h\] and to find h i substitute
\[\frac{3}{2}\] for x to get the answer

Answer 17

so i would get
\[y=2(\frac{3}{2})^2-6\times \frac{3}{2}+9\]

Answer 18

making the vertex
\[(\frac{3}{2},\frac{9}{2})\]
and
\[y=2(x-3)^2+\frac{9}{2}\]

Answer 19

that way i don't have to keep track of what i added and subtracted while completing the square