Factoring Quadratic Equation?

**NotSoBasic** · June 20th 2009, 02:21 PM

Just re-learning the quadratic equation.

I understand it until the end where the example in my book shows when solving:

$x= \dfrac {-4\pm2\sqrt{6}}{4}$

Then:

$x= -1- \dfrac {\sqrt{6}}{2}$ or, $x=-1+\dfrac {\sqrt{6}}{2}$

Not quite sure how they factored, it does not explain. I tried, but kept coming up with.

$x= -2 \pm \dfrac {\sqrt{6}}{2}$

Somebody care to explain how it's done properly..

**stapel** · June 20th 2009, 02:40 PM

What is the value of four divided by four?

**NotSoBasic** · June 20th 2009, 02:44 PM

Originally Posted by stapel

What is the value of four divided by four?

It's 1.

Wouldn't that cancel out the denominator?

leaving,

$x=-1\pm2\sqrt{6}$

**e^(i*pi)** · June 20th 2009, 03:03 PM

Originally Posted by NotSoBasic

It's 1.

Wouldn't that cancel out the denominator?

leaving,

$x=-1\pm2\sqrt{6}$

$<br /> x= \dfrac {-4\pm2\sqrt{6}}{4} = \frac{-4}{4} \pm \frac{2\sqrt6}{4}<br />$

Now what does 4/4 and 2/4 equal

**NotSoBasic** · June 20th 2009, 05:12 PM

Ah, OK. That makes sense.. =D

Is that step called something where you separate the values, I would like to research more on that.

**Stroodle** · June 20th 2009, 05:34 PM

It's called reducing, or simplifying, fractions.

**VonNemo19** · June 20th 2009, 05:59 PM

Originally Posted by NotSoBasic

Ah, OK. That makes sense.. =D

Is that step called something where you separate the values, I would like to research more on that.

I'm not sure about what it's called, but I can prove it.

$\frac{a+b}{c}=\frac{1}{c}\cdot(a+b)=\frac{a}{c}+\f rac{b}{c}$

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