# Thread: Completing the Square problem solving

1. ## Completing the Square problem solving

$\displaystyle 3p^2-12p+2=0$
$\displaystyle p^2-4p+\frac{2}{3}=0$
Completing the square
$\displaystyle (\frac{4}{2})^2=4$ and $\displaystyle \frac{2}{3}+x=4$ so add $\displaystyle \frac{10}{3}$ to each side.
$\displaystyle p^2-4p+4=\frac{10}{3}$ which is $\displaystyle (p-2)^2=\frac{10}{3}$
I end up with $\displaystyle p=2 \pm \sqrt{\frac{10}{3}}$

The book says $\displaystyle \frac{6 \pm \sqrt{30}}{3}$

Whats up with that?

2. Originally Posted by bkbowser
$\displaystyle 3p^2-12p+2=0$
$\displaystyle p^2-4p+\frac{2}{3}=0$
Completing the square
$\displaystyle (\frac{4}{2})^2=4$ and $\displaystyle \frac{2}{3}+x=4$ so add $\displaystyle \frac{10}{3}$ to each side.
$\displaystyle p^2-4p+4=\frac{10}{3}$ which is $\displaystyle (p-2)^2=\frac{10}{3}$
I end up with $\displaystyle p=2 \pm \sqrt{\frac{10}{3}}$

The book says $\displaystyle \frac{6 \pm \sqrt{30}}{3}$

Whats up with that?

$\displaystyle \frac{6 \pm \sqrt{30}}{3} = 2 \pm \frac{\sqrt{30}}{3} = 2 \pm \frac{\sqrt{3}\sqrt{10}}{\sqrt{3}\sqrt{3}} = 2 \pm \sqrt{\frac{10}{3}}$

3. Originally Posted by Defunkt

$\displaystyle \frac{6 \pm \sqrt{30}}{3} = 2 \pm \frac{\sqrt{30}}{3} = 2 \pm \frac{\sqrt{3}\sqrt{10}}{\sqrt{3}\sqrt{3}} = 2 \pm \sqrt{\frac{10}{3}}$
OK why does the $\displaystyle 2 \pm \frac{\sqrt{3}\sqrt{10}}{\sqrt{3}\sqrt{3}}$ section work like that? I get the nominator portion but not the denominator.

4. $\displaystyle \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$

$\displaystyle b = \sqrt{b} \cdot \sqrt{b}$

Can you see why now?

5. Originally Posted by Defunkt
$\displaystyle \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$

$\displaystyle b = \sqrt{b} \cdot \sqrt{b}$

Can you see why now?
Ya, since three can be written as a product of the square root of 3 and the square root of three.

Thats pretty cool by the way, never really thought about doing that.