Question

can someone please explain:Complete the square to write the equation in
the standard form f (x) = a(x − h)^2 + k . For quadratic functions please?

Answer 1

I don't understanding completing the square part for 3x^2+18x+15

Answer 2

it is time to get familiar with this shortcut:

\[(a + b)^2 = a^2 + 2ab + b^2\]and\[(a - b)^2 = a^2 - 2ab + b^2\]

Answer 3

ok

Answer 4

lets take the 3 out of you polynomial to make it a bit easier to look at:

\[3(x^2 + 6x + 5)\]
lets just look at

\[x^2 + 6x + 5\]for now

completing the square is applying this shortcut. our equation looks like:
\[(x + b)^2 = x^2 + 2bx + b^2\]
since we know that 6x represents 2xb, we can find the b value that would complete the square. this value is b = 3

\[(x+3)^2 = x^2 + 6x + 9\]
so our original equation (x^2 + 6x + 5) an be re-written as:\[(x+3)^2 - 4\]

we can now multiply back everything by 3:

\[3(x + 3)^2 - 12\]
is the re-written form

Answer 5

oops, another way to look at it is:
\[x^2 + bx = (x + \frac{ b }{ 2 })^2 - \left( \frac{ b }{ 2 } \right)^2\]

Answer 6

let me know if you have questions. i can solve more if you have some ^_^

Answer 7

where did you get the 4 in : (x+3)^2-4?

Answer 8

and we need to take that 3 out of 3x^2 and put it in 3(x^2+6x+5) and because of that we take (6/2)^2 which is 3 and that's what we are applying to the b.

Answer 9

soory - (6/2)2 is 9

Answer 10

i'll use the "other way to look at it"

completing the square on x^2 + 6x gives bring us to the same value because it is simply another way of writing it

\[(x + \frac{ 6 }{ 2 })^2 - \left( \frac{ 6 }{ 2 } \right)^2 = (x + 3)^2 - 9 = x^2 + 6x + 9 - 9 = x^2 + 6x\]
as expected.

so our equation if we include the +5:
\[(x+3)^2 - 9 + 5 = (x+3)^2 - 4\]

Answer 11

ohhhh ok for some reason I like the other way . I am going to try it to a different one and ill post it here for practice

Answer 12

sounds good :)

Answer 13

k im going to solve: 5x^2 −10x + 8

Answer 14

let me know when i should solve it ^_^

id suggest:

1st step: factor out the 5 [the x^2 coefficient] even though it isn't always neat for all terms
2nd step: complete the square on x^2 - 2x
3rd step: add or subtract a constant to adjust it such that c in ax^2 + bx + c = 8

Answer 15

5(x-2x+4) ? the 8 throws me off from the 5 and -10

Answer 16

5 X 2 is 10 but 5 times nothing is 8

Answer 17

but if I use the other way - x^2+bx=(x+b/2)^2-(b/2)^2
= x^2+(-5)x=(x+(-5) - (-5)

I have a feeling I did that very wrong ... =(

Answer 18

solve?

Answer 19

sorry got phone call

\[5(x^2 - 2x) + 8\]\[5[(x-1)^2 -1] + 8 = 5(x-1)^2 + 3\]

Answer 20

did the 3 come from 8-5? because (x-b/2)^2 and (-10/2)^2 = -5?

Answer 21

sorry im like the worst person at math and I swear it takes me a million times longer to get things and these are hard for me but I have to get them down ... thank you for your patience

Answer 22

why did you put -1]?

Answer 23

yes the three is from 8 - 5

and the -1 was used to "complete the square"

\[(x-1)^2 - 1 = x^2 - 2x\]

expanding (x -1)^2, using shortcut or whatever method, gives x^2 - 2x + 1.

we add -1 to be left with x^2 - 2x.

x^2 - 2x = (x - 1)^2 - 1

Answer 24

the square is just the (x-1)^2 part. but completing the square is adding or subtracting whatever you need from the square to return to your original equation

Answer 25

isn't -1 x -1 = 1? not -1?

Answer 26

yes. so we have

x^2 - 2x + 1 - 1 = x^2 - 2x

Answer 27

(x-1)^2 include the + 1. we're subtracting 1 to get only x^2 - 2x

Answer 28

is it okay if I start over with a different equation if you have time?

Answer 29

yes.

sorry, i keep forgetting about OpenStudy because a friend of mine keeps calling me.

if you post a question i'll solve it :)