Complete the Square

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January 4th 2009, 12:57 PM
magentarita

Complete the Square

Complete the square for 3x^2 - 7x + 12 = 0

MY WORK:

I know that the coefficient of the x^2 term must be 1 in order to complete the square.

I divided everything by 3 and came up with this:

x^2 -(7x/3) + 4 = 0

I then subtracted 4 from both sides and got this:

x^2 - (7x/3) = -4

I then took 1/2 of -7/3, which is 14/9, and squared that to add to both sides and got this:

x^2 - (7x/3) + (196/81) = -4 + (196/81)

Where do I go from there?
January 4th 2009, 01:10 PM
Simplicity

Quote:

Originally Posted by magentarita

Complete the square for 3x^2 - 7x + 12 = 0

MY WORK:

I know that the coefficient of the x^2 term must be 1 in order to complete the square.

I divided everything by 3 and came up with this:

x^2 -(7x/3) + 4 = 0

I then subtracted 4 from both sides and got this:

x^2 - (7x/3) = -4

I then took 1/2 of -7/3, which is 14/9, and squared that to add to both sides and got this:

x^2 - (7x/3) + (196/81) = -4 + (196/81)

Where do I go from there?

Don't subract 4 at that point. First complete the square at that point.

$\left(x+\frac{7}{6}\right)^2-\left(\frac{7}{6}\right)^2 + 4=0$

Then solve for $x$ using algebra techniques.

EDIT: You will get a complex answer.
January 4th 2009, 01:14 PM
galactus

When completing the square, remember the general form. Then all you have to do is plug in your a, b, c.

$a\left(x+\frac{b}{2a}\right)^{2}+\underbrace{c-\frac{b^{2}}{4a}}_{\text{constant}}$

Take yours

a=3, b=-7, c=12

$3\left(x+\frac{-7}{2(3)}\right)^{2}+12-\frac{(-7)^{2}}{4(3)}$

$3\left(x-\frac{7}{6}\right)^{2}+\frac{95}{12}$
January 5th 2009, 09:56 PM
magentarita

7/6....

Quote:

Originally Posted by galactus

When completing the square, remember the general form. Then all you have to do is plug in your a, b, c.

$a\left(x+\frac{b}{2a}\right)^{2}+\underbrace{c-\frac{b^{2}}{4a}}_{\text{constant}}$

Take yours

a=3, b=-7, c=12

$3\left(x+\frac{-7}{2(3)}\right)^{2}+12-\frac{(-7)^{2}}{4(3)}$

$3\left(x-\frac{7}{6}\right)^{2}+\frac{95}{12}$

Where did the fraction 7/6 come from?
January 5th 2009, 10:27 PM
mathaddict

Re :

$f(x)=3x^2-7x+12$
$f(x)=3(x^2-7/3x+4)$
$f(x)=3[(x-7/6)^2+4-(7/6)^2]$

Do you see now ?
January 6th 2009, 03:33 AM
HallsofIvy

Quote:

Originally Posted by magentarita

Complete the square for 3x^2 - 7x + 12 = 0

MY WORK:

I know that the coefficient of the x^2 term must be 1 in order to complete the square.

I divided everything by 3 and came up with this:

x^2 -(7x/3) + 4 = 0

I then subtracted 4 from both sides and got this:

x^2 - (7x/3) = -4

I then took 1/2 of -7/3, which is 14/9,

1/2 of -7/3 is NOT 14/9, it is -7/6.

Quote:

and squared that to add to both sides and got this:

Quote:

x^2 - (7x/3) + (196/81) = -4 + (196/81)

Where do I go from there?
January 6th 2009, 06:35 AM
magentarita

yes...

Quote:

Originally Posted by mathaddict

$f(x)=3x^2-7x+12$
$f(x)=3(x^2-7/3x+4)$
$f(x)=3[(x-7/6)^2+4-(7/6)^2]$

Do you see now ?

Yes, I do....Thanks
January 6th 2009, 06:37 AM
magentarita

I see...

Quote:

Originally Posted by HallsofIvy

1/2 of -7/3 is NOT 14/9, it is -7/6.

I see what I did wrong.

I divided -7/3 by 1/2 and got -14/9.

I had to multiply -7/3 by 1/2 to get -7/6.

I got it now. Thanks.
January 6th 2009, 08:35 AM
mr fantastic

Quote:

Originally Posted by magentarita

I see what I did wrong.

I divided -7/3 by 1/2 and got -14/9.

I had to multiply -7/3 by 1/2 to get -7/6.

I got it now. Thanks.

Note: $- \frac{7}{3} \div \frac{1}{2} \neq -\frac{14}{9}$ .Rather, it's $- \frac{14}{6}$ .
January 6th 2009, 10:35 AM
craig

Quote:

Originally Posted by mr fantastic

Note: $- \frac{7}{3} \div \frac{1}{2} \neq -\frac{14}{9}$ .Rather, it's $- \frac{14}{6}$ .

Isn't it $- \frac{14}{3}$ ?
Sorry was just flicking through the forum ;)
January 6th 2009, 01:03 PM
mr fantastic

Quote:

Originally Posted by craig

Isn't it $- \frac{14}{3}$ ?
Sorry was just flicking through the forum ;)

Well, on Earth it certainly is. But not on a backward planet named toidi, which is where I was visiting at the time.
January 6th 2009, 01:04 PM
craig

Quote:

Originally Posted by mr fantastic

Well, on Earth it certainly is. But not on a backward planet named toidi, which is where I was visiting at the time.

Glad to see it happens to the best of us at times ;)
January 6th 2009, 08:53 PM
magentarita

ok...

-7/6 divided by 1/2 = -7/6 X 2/1 = -14/6

I spend so much time with high school level math that primary school math now seems distant.(Happy)