Completing the square

Question

Can someone explain what happens from step 6 forward, when solving 3x^2 – 12x – 7 = 0 by completing the square. How does radical of 19/3 turn into radical 57/ 3?

Answer 1

Multiply numerator and denominator by $\sqrt{3}$ and you get

$$\sqrt{\frac{19}{3}}=\frac{\sqrt{57}}{3}$$

Answer 2

This has nothing to do with completing the square. Your confusion is in "rationalizing the demoninator" (usually a square root--but that need not be the case).

In your case you have $\frac{\sqrt{19}}{\sqrt{3}}$. We would "like" to get rid of the divide by an irrational number (although this is an unnecessary step introduced largely for arbitrary reasons by high school math teachers--but it does teach you some properties of fractional powers). To get rid of it, we simply multiply by a special form of "1":

$$ \frac{\sqrt{19}}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3*19}}{\sqrt{3*3}} = \frac{\sqrt{57}}{3} $$

That simplification has nothing to do with completing the square--it's just a simplification which would have been just as correct to leave it out and keep the answer as:

$$ 2 \pm \sqrt{\frac{19}{3}} $$

Another Example of Rationalizing the denominator:

I don't like it when you see nothing but square roots. Here's another example:

$$ \frac{1}{\sqrt[3]{2}} $$

We don't multiply by $\frac{\sqrt[3]{2}}{\sqrt[3]{2}}$ because this doesn't rationalize the denominator...we need to get a whole value for $2$:

$$ \frac{1}{\sqrt[3]{2}} * \frac{\sqrt[3]{2^2}}{\sqrt[3]{2^2}} = \frac{\sqrt[3]{4}}{\sqrt[3]{2^3}} = \frac{\sqrt[3]{4}}{2} $$

Had it been $\frac{1}{\sqrt[3]{4}}$ then we could have either done $\frac{1}{\sqrt[3]{4}}*\frac{\sqrt[3]{16}}{\sqrt[3]{16}} = \frac{\sqrt[3]{2^4}}{4} = \frac{\sqrt[3]{2^3*2}}{4} = \frac{2*\sqrt[3]{2}}{4} = \frac{\sqrt[3]{2}}{2}$ or we could already recognize that $\sqrt[3]{4} = \sqrt[3]{2^2}$ which is only lacking a $\sqrt[3]{2}$:

$$ \frac{1}{\sqrt[3]{4}}*\frac{\sqrt[3]{2}}{\sqrt[3]{2}} = \frac{\sqrt[3]{2}}{2} $$