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I need to prove that
ax^2 + bx + c = a(x-h)^2 + k
This is what I have so far

a(x-h)^2 + k = 0

a(x-h)(x-h) + k = 0

ax^2 - xh - xh + h^2 + k = 0

ax^2 - 2xh + h^2 + k = 0

Do you have any ideas on how to change
a(x-h)^2 + k
into the format
ax^2 + bx + c

Anything would be appreciated!

2007-05-25 16:20:55 · 6 answers · asked by Anonymous in Science & Mathematics Mathematics

6 answers

Start with ax²+bx+c. Then factor a:

a [x² + (b/a)x + (c/a)]

Then, complete the square the same way you would do when deriving the quadratic formula:

= a [x² + (b/a)x + b²/4a² -b²/4a² + (c/a)]
= a [(x+b/2a)² - b²/4a² + c/a]
= a (x+b/2a)² -b²/4a + c
and basically, you're done. For an equation of the form

a(x-h)² + k

your equivalent coefficients are

h = -b/2a
k = -b²/4a + c

to put the equation into the desired form.

(This answer was updated Saturday 1306 PDT to correct the botched algebra at the end. Sorry, my bad.)

Good luck, work hard, and stay away from drugs.

2007-05-25 16:25:05 · answer #1 · answered by MikeyZ 3 · 1 0

I seems like you are trying to reinvent the wheel. This looks alot like the basic "sum of squares" routine from which the quadratic formula can be derived. I'd start with the ax2+bx+c and proceed forward rather than proceed back. First you can multiply by a to get
a^2x^2+ ba x + ca
Without doing a lot of arm waving, there is a constant d for which a^2x^2 + ba x + d^2 is a perfect square. Then we have
a^2 x^2 + ba x + d^2 + ca -d^2. Then
(ax+d)^2 + ca - d^2
Now, all we have to do is set d= -ah
and (ax-ah)^2 + ca-d^2
setting ca-d^2 = k completes the exercise.

2007-05-25 16:41:02 · answer #2 · answered by cattbarf 7 · 0 0

Use "foil" on the (x-h)^2 part.

foil stands for first, outside, inside, last.

rewriting with the a: a(x-h)(x-h)

Multiply the first ones .... x^2
Multiply the outside ones .... -xh
Multiply the inside ones .... another -xh
Mulitply the last ones .... h^2

Add them all together x^2 -xh -xh +h^2

Simplify and remember to include the a

a(x^2 - 2xh + h^2)

2007-05-25 16:27:14 · answer #3 · answered by BP 7 · 0 0

Do you remember the order of operations? First square (x-h) to get x^2 - 2hx +h^2. Then multiply the result by a to get
ax^2 + (-2ahx) +ah^2
let -2ah = b
let ah^2 =c and the answer becomes
ax^2 + bX + c
The tricky part here is remembering tha (a-h) is squared first and then multiply by a. Also b and c are constants and can be set equal to -2ah and ah^2 (which are also constants).

2007-05-25 16:42:54 · answer #4 · answered by skipper 7 · 0 0

You're almost on the right track.

a(x-h)^2+k = ax^2-2xh+h^2+k

let a=a
let b=-2h
let c=h^2+k

all of which are constants so you can do this.

then you have

ax^2-2xh+h^2+k = ax^2 + (-2h)(x) + (h^2+k) = a^2 + bx + c

2007-05-25 16:30:36 · answer #5 · answered by The Accountant 2 · 0 0

a(x-h)^2 = ax^2 -2axh +ah^2
add k to put it into the right form
a(x-h)^2 = ax^2 -2axh +ah^2 +k = ax^2 + bx + c

so b=-2ah and c = ah^2+k

2007-05-25 16:32:42 · answer #6 · answered by davec996 4 · 0 0

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