Question

Complete the Square:

Answer 1

\[2x^2+8x-7=-2\]

Answer 2

break it down to a simply problem

Answer 3

then do the pemdas stage

Answer 4

would you start by adding seven to both sides?

Answer 5

\[2x^2+8x-7=-2\]

Answer 6

You could do that, but I don't see how that step would help you.
This is an equation, namely, a quadratic equation, and you'll want all terms on the left side.

Answer 7

If you want to move the -2, fine. Add 2 to both sides of this equation.

Answer 8

so 2x^2+8x-5=0

Answer 9

Yes. Now make things easier for yourself by factoring 2 out of the first 2 terms. Use parentheses. Leave the -5 alone.

Answer 10

2(x^2+4x)-5=0?

Answer 11

Yes, very good. Now focus on completing the square of x^2+4x.

Answer 12

1) Take half of the coefficient of x (which is 4).
2) Square your result.
3) Add this result to the first 2 terms, and then subtract this result from the left 3 terms.
Show your work.

Answer 13

I'm confused. what terms do i have to add or subtract it from? i get the whole (b/2)^2 , I'm not sure on where to add this too

Answer 14

b is 4.
Half of b is 2.
Square this result. It's 4.
Add this 4, and then subtract this 4: x^2+4x +4 -4

Answer 15

x^2+4x+4 is the square of what binomial?

Write your answer in the form ( )^2-4.

Answer 16

(x+2)^2-4

Answer 17

Yes. Now, let's put this problem back together:

2x^2+8x-5=0
2(x^2+4x)-5 = 0
Replace (x^2+4x) with ( (x+2)^2 -4 ).

Show your work.

Answer 18

so it'd be 2(x+2)^2-9=0?

Answer 19

Here's what I would do:

2 [(x^2+4x+4 - 4)] - 5 = 0

2 [ (x+2)^2 - 4] - 5 = 0

2(x+2)^2 - 8 -5 = 0

Please go thru this and decide whether you can accept it or not.

Answer 20

okay i think i got it. thank you

Answer 21

My pleasure! Bye for now.