1. 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?

2. 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.

3. 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

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

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

4. 7/6....

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?

5. 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 ?

6. 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.

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?

7. yes...

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

Do you see now ?
Yes, I do....Thanks

8. I see...

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.

9. 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 $- \frac{14}{6}$.

10. 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

11. 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.

12. 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

13. 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.