# Operating and rationalizing surds

can someone show me the methodology for dealing with such operations of surds as these:

The question is to express as a single fraction with a rational denominator.

I would post my working out, but it has absolutely no link to the answer, and it's because I have never been taught how to solve these types of questions before. If you could explain to me the methodology for solving these and how you did it, it would be very much helpful. Thanks

$$\frac{\sqrt{2}}{\sqrt{2}-\sqrt{3}} -\frac{3}{\sqrt{2}+\sqrt{3}}$$

$$\frac{\sqrt{2}(\sqrt{2}+\sqrt{3}) - 3\sqrt{2} + 3\sqrt{3}}{(\sqrt{2}-\sqrt{3}) (\sqrt{2}+\sqrt{3})}$$

Apply $(a+b)(a-b) = a^2 - b^2$

I think you can take it from here?

• Can you please explain what you did? Commented Mar 11, 2014 at 10:00
• @Brett $$\frac{a}{b} - \frac{c}{d} = \frac{ad-bc}{bd}$$ You know how to take the LCM?
– Guy
Commented Mar 11, 2014 at 10:01
• Yes, but I don't understand what you did in the answer. Did you multiply everything by the same expression? It seems that you just combined the denominator, and I'm not sure what you did with the numerator. Commented Mar 11, 2014 at 10:03
• @Brett No. I multiplied the $\sqrt{2}$ in the first fraction by $(\sqrt{2}+\sqrt{3})$,i.e the denominator of second fraction. Like you see $ad$ in my example. Similarly I multiplied $3$ by $\sqrt{2} - \sqrt{3}$. Get it now? And the denominator is $(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})$ which simplifies to $2-3 = -1$
– Guy
Commented Mar 11, 2014 at 10:06
• OK, thanks I see it clearly now. Regarding the rule you posted in wherein a/b - c/d = ad-bc / bd, does this rule apply to all fractions, even when you add? eg a/b + c/d = ad + bc / bd ? Commented Mar 11, 2014 at 10:10

To the comment, you can see this by noticing, for $b,d \not = 0$

\begin{align*} \frac{a}{b} \pm \frac{c}{d} = \frac{a}{b}\left(\frac{d}{d}\right) \pm \frac{c}{d}\left(\frac{b}{b}\right) = \frac{ ad \pm bc}{bd} \end{align*}

I think the best way is to rationalize the denominator of each term separately.

$$\frac{\sqrt2}{\sqrt2-\sqrt3}=\frac{\sqrt2(\sqrt2+\sqrt3)}{(\sqrt2-\sqrt3)(\sqrt2+\sqrt3)}=\frac{2+\sqrt6}{2-3}$$

Again, $$\frac3{\sqrt2+\sqrt3}=\frac{3(\sqrt2-\sqrt3)}{(\sqrt2+\sqrt3)(\sqrt2-\sqrt3)}=\frac{3\sqrt2-3\sqrt3}{2-3}$$