How do I express 14/(3-√2) in the form of b+c√d?

Question

I don't really understand the method to express a surd. Can anyone help me with this question? I really want more ideas on how to solve it, maybe there is one I may understand well. TY

Answer 1

It's called rationalizing the denominator. This website actually has an example similar to your problem.

$$ \begin{align} \frac{14}{3-\sqrt{2}} &= \frac{14 (3 + \sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}\\ &= \frac{42 + 14\sqrt{2}}{7} \\ &= 6 +2\sqrt{2} \end{align} $$

Answer 2

$\Bbb Q[\sqrt{2}]$ is a field (with the usual definitions of $+$ and $\cdot$).

What that means is that you can add, subtract, multiply, and divide two numbers of the type $a+b\sqrt{2}$ and you'll get another number of that type.

For your specific problem, first notice that $(c+d\sqrt{2})(c-d\sqrt{2}) = c^2 -2d^2$ is just a rational number (it is also of the form $a+b\sqrt{2}$, where $b=0$).

So then we have the idea that if we want to get a number of the form $c+d\sqrt{2}$ out of the denominator, we can just multiply it by $c-d\sqrt{2}$ so that it becomes a regular rational number.

Let's try it:

$$\frac{14}{3-\sqrt{2}} = \frac{14}{3-\sqrt{2}}(1) = \frac{14}{3-\sqrt{2}}\left(\frac{3+\sqrt{2}}{3+\sqrt{2}}\right) = \frac{42+14\sqrt{2}}{7} = 6 + 2\sqrt{2}$$