Expert: Sherman D. - 9/22/2006

Question
"at the point you have to divide by half, then square it, and add the result to both sides."

Right. You Have to do that. But what does the quadratic formula have to do with completing the square? My book doesn't say anything bout that...
I looked it up and it looked to me like two diff formulas.... I'm not trying to be rude or anything, but i'm just confused. And it's not that i don't believe you. I'm not good at any kind of math. Period.

also this is what i found:

http://www.classbrain.com/artteensb/publish/completing_the_square_quadratic_form

Also, i'm sorry to bother u.
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Followup To

Question -
I think I understand now... but one last thing... when you take the half of the 3/2, ain't ya supposed to add that 9/16 to both sides, not multiply it. Shouldn't it be 9/16 + 9/16? I'm not trying to be a know it all or anything... if i knew it all i wouldn't be asking you, LOL. But really, i'm just checking...
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Followup To

Question -
Hi, I really need help. I'm stuck on this problem and i can't get out. No Really. I need help.

It says to Solve by completing the square. My book doesn't give me but two pitiful examples.

I did some work, not sure if it's right...

2x^2 - 3x + 1 = 0 (Original Problem)

2x^2 - 3x + 1 - 1 = 0 - 1

2x^2 - 3x = -1

2x^2/2 - 3x/2 = -1/2

x^2 - 3x/2 = -1/2

(That's where I'm stuck, If any of that is correct. I hope you can help. I heard you guys were good at this stuff. Thanks in advance.)

Answer -
2x^2 - 3x + 1 = 0
2x^2 - 3x = -1
x^2 - (3/2)x = (-1/2)
x^2 - (3/2)x + (9/16) = (1/16)
(x - (3/4))^2 = (1/16)
x - (3/4) = sqrt(1/16)
x - (3/4) = (±1/4)
x = (3/4) ± (1/4)
x = (4/4) or (2/4)

x = 1 or (1/2)

Answer -
at the point you have to divide by half, then square it, and add the result to both sides.

here is the quadratic formula, which is the conclusion of completing the square.

x = (-b ± sqrt(b^2 - 4ac))/(2a)

if you don't believe me, just look up "completing the square and the quadratic formula"

Answer
"the QUADRATIC FORMULA is the CONSLUSION to the COMPLETING THE SQUARE method "

if you put your coeficients into the quadratic formula, you will get x = 1 or (1/2)

ax^2 + bx + c = 0

2x^2 - 3x + 1 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-3) ± sqrt((-3)^2 - 4(2)(1)))/(2(2))
x = (3 ± sqrt(9 - 8))/4
x = (3 ± sqrt(1))/4
x = (3 ± 1)/4
x = (3 + 1)/4 or (3 - 1)/4
x = 4/4 or (2/4)
x = 1 or (1/2)

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ax^2 + bx + c = 0

ax^2 + bx = -c
x^2 + (b/a)x = (-c/a)

find half of (b/a), which is (b/(2a)) square it and add to both sides

x^2 + (b/a)x + ((b^2)/(2a^2)) = ((b^2)/(2a^2)) - (c/a)

which is the same as saying

x^2 + (b/a)x + ((b^2)/(4a^2)) = ((b^2 - 4ac)/(4a^2))

factor the left side into a perfect square

(x + (b/(2a)))^2 = ((b^2 - 4ac))/(4a^2))

sqrt both sides

x + (b/(2a)) = sqrt((b^2 - 4ac)/(4a^2))

same as saying

x + (b/(2a) = (sqrt(b^2 - 4ac))/(2a)

subtract (b/(2a)) from both sides

x = (-b/(2a)) ± (sqrt(b^2 - 4ac))/(2a))

which you can combine to

x = (-b ± sqrt(b^2 - 4ac))/(2a) which is the quadratic formula.

all i did was, i used the completing the square method, but instead of using numbers, i used letters in the place of those numbers, since the numbers can be value, but the quadratic formula applies to any number that a, b, or c might be.

the reason your book probably doesn't have it, is because you probably won't learn it yet, and the reason they don't give it to you is because using the completing the square method is a way of learning, and the quadratic formula is a short cut to those who already know how to do the completing the square method. All the quadratic formula does is makes it easier for you to find the answer.

I even use the quadratic formula to help me factor, for ex:

(ax - b)(cx - d)

as we know, the x values are 1 and (1/2), which is the same as (1/1) and (1/2),

a and c are the denominators and b and d are the numerators.

let me show you what i mean.

(1/1) = (b/a)
(1/2) = (2/1)

so just plug those in, and you get (x - 1)(2x - 1). The reason why i put (x - 1)(2x - 1), instead of (x + 1)(2x + 1), is because you need (x - 1) or (2x - 1) to equal zero, and if you plug in the values for x that you know from above, they will give you zero.