Completing The Square

**liveboon** · February 15th 2010, 06:40 PM

Do people know how to complete the square

For Example:
3x2 - 15x + 18 = 0

(Note: 3x2 means 3x squared)

**Mathless** · February 15th 2010, 06:44 PM

pull out the common factor of 3, which leaves you:

3(x^2 - 5x + 6) = 0 ----> 3(x-3)(x-2) = 0

x= 3 and x=2

**liveboon** · February 15th 2010, 06:45 PM

OMG!! Mathless i absolutely love u!!!!

**Prove It** · February 15th 2010, 06:45 PM

Originally Posted by liveboon

Do people know how to complete the square

For Example:
3x2 - 15x + 18 = 0

(Note: 3x2 means 3x squared)

$3x^2 - 15x + 18 = 0$

$3(x^2 - 5x + 6) = 0$

$3\left[x^2 - 5x + \left(-\frac{5}{2}\right)^2 - \left(-\frac{5}{2}\right)^2 + 6\right] = 0$

$3\left[\left(x - \frac{5}{2}\right)^2 - \frac{25}{4} + \frac{24}{4}\right] = 0$

$3\left[\left(x - \frac{5}{2}\right)^2 - \frac{1}{4}\right] = 0$

$3\left(x - \frac{5}{2}\right)^2 - \frac{3}{4} = 0$ .

**Prove It** · February 15th 2010, 06:47 PM

P.S. Someone please lock this thread - there's no need for outbursts like that!

**tonio** · February 15th 2010, 06:52 PM

Originally Posted by liveboon

Do people know how to complete the square

For Example:
3x2 - 15x + 18 = 0

(Note: 3x2 means 3x squared)

In general, $ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2 +c-\frac{b^2}{4a}$ , and perhaps you can recognize the term $c-\frac{b^2}{4a}=-\frac{\Delta}{4a}\,,\,\,\Delta=$ the quadratic's discriminant.

So yes, I think some people know how to complete the square.

Tonio

**tonio** · February 15th 2010, 06:55 PM

Originally Posted by Mathless

pull out the common factor of 3, which leaves you:

3(x^2 - 5x + 6) = 0 ----> 3(x-3)(x-2) = 0

x= 3 and x=2

I'm afraid this is not what the OP asked for, though she/he seems to be unaware of it. What you did here is to factor the quadratic, which is very useful to calculate its roots, for example, but it's not completing the square.

Tonio

**Chris L T521** · February 16th 2010, 07:29 AM

I've seen enough happen here in this thread. Now that the question has been answer, there is no need to continue posting here.

Thread closed.

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