Question

saifoo please help...complete the square
-2((x^2)+3x-4)...im not sure how to factor after you do the (b/2)^2

Answer 1

Sorry to say i suck @ completing the square.

Sat & others rules at that.

Answer 2

that is for sure!! lol

Answer 3

can you help me satellite73? =]

Answer 4

is this
\[-2(x^2+3x-4)\]?

Answer 5

i wanna learn that too. :D

Answer 6

yes
actually it started off as a transformation and then simplified to this and then it was said to find answer by completing the square

Answer 7

(if that gets into my head) :[

Answer 8

ok why don't we start with
\[-2(x^2+3x)+8\] instead.

Answer 9

http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e8e565a0b8b543d4218f9fd

Answer 10

then half of 3 is 3/2, and 3/2 squared is 9/4 so we write
\[-2(x+3x+\frac{9}{4})+8+2\times \frac{9}{4}\]

Answer 11

x^2 + 3x - 4 is not a perfect square, but
x^2 + 3x + (3/2)^2 is a perfect square
3/2^2 = 9/4, which is 2.25
In order to add 2.25, you have to also subtract it so that you don't change the value:
-2((x^2)+3x + 2.25 - 2.25 -4)
Then simplify:
-2((x+3/2)^2 - 2.25-4)
-2((x+3/2)^2 - 6.25)

Answer 12

and the reason the last part is
\[+2\times \frac{9}{4}\] is because you are actually subtracting that in the first term by the distributive law

Answer 13

the original equation was f(x)=-2x^2-6x+8

Answer 14

then you have
\[-2(x+\frac{3}{2})^2+8+\frac{9}{2}\] or
\[-2(x+\frac{3}{2})^2+\frac{25}{2}\]

Answer 15

ok well then i think we have the right answer

Answer 16

-2(x^2 + 3x + 9/4 - 4 - 9/4) = 0
x^2 + 3x + 9/4 = 4 + 9/4
(x + 3/2)^2 = 4 + 9/4
4(x+3/2)^2 = 16 + 9 = 25

2(x+3/2) = 5 or 2(x+3/2) = -5

2x + 3 = 5 or 2x + 3 = -5

x = 1 or x = -4

Answer 17

i will write it again step by step
\[f(x)=-2x^2-6x+8\]
\[f(x)=-2(x^2+3x)+8\]
\[f(x)=-2(x^2+3x+\frac{9}{4})+8+2\times \frac{9}{4}\]
\[f(x)=-2(x+\frac{3}{2})^2+\frac{25}{2}\] and that is it

Answer 18

ok ummm once again how did you get the 2x(9/4) thing...like why?
and how did you factor out the (3/2)

Answer 19

you did ask to complete the square, not to set it equal to zero and solve correct?

Answer 20

yea i needed the complete the square

Answer 21

ok first where did the 3/2 come from.
the middle term is
\[3x\] so i need to take half of 3, and that is
\[\frac{3}{2}\]

Answer 22

so once i "complete the square " my answer will look like
\[(x+\text{ something })^2\] and that "something" has to be half of the coefficient of the middle term. if you multiply out you will see why

Answer 23

\[x^2+3x+9/4\]is a perfect square created by completing the square, which simplifies to \[(x+\frac{3}{2})^2\]satellite got the 2x(9/4) thingy by taking (-9/4) outside of the expression in the parentheses

Answer 24

now when replace
\[x^2+3x\] by
\[(x+\frac{3}{2})^2\] they are not the same thing

Answer 25

because
\[(x+\frac{3}{2})^2=x^2+3x+\frac{9}{4}\] so i have actually added
\[\frac{9}{4}\]

Answer 26

but in this case we had
\[-2(x^2+3x)\] so when i replace that by
\[-2(x+\frac{3}{2})^2\] i did not actually add
\[\frac{9}{4}\] but in fact i subtracted
\[2\times \frac{9}{4}\] because of that pesky -2 out front

Answer 27

so in order to keep the function the same, i had to add it on back at the end. that is why i wrote
\[+2\frac{9}{4}\] at the end.
there is an easier way to do this however

Answer 28

what is the easier way?

Answer 29

nvm i got it!!! =] thnx! =]