Simplifying fractions

Released on 12. 9. 2024 in category Výrazy; Author: Jakub Vojáček; Read count: 71

When performing an operation called fraction rationalization, we eliminate square roots in the denominator while preserving the value of the fraction.

As I already hinted, we will learn to remove the square root from the denominator. Essentially, it's a very simple task, but there are a few things to keep in mind.

That was the first thing. The second thing is:

Let's try this with an example. Try to rationalize the fraction $\frac{1}{\sqrt{2}}$ . The denominator is √2. Therefore, we multiply the entire fraction by $\frac{\sqrt{2}}{\sqrt{2}}$ .

You see, it was easy;-) Let's try a few more examples:

Rationalize the fraction:

In the previous example, we multiplied the entire fraction by the value of the denominator, but in this case, it would be unnecessary. Of course, we could multiply the entire fraction by $\frac{2*\sqrt{2}}{2*\sqrt{2}}$ , but an easier solution is to multiply only by $\frac{\sqrt{2}}{\sqrt{2}}$ . So the whole thing would look like this:

We have a rationalized fraction, but we can still simplify 4 and get the result:

These were easier examples. Slightly more problems arise when there are more terms in the denominator. For example, try rationalizing the fraction $\frac{1}{1+\sqrt{x}}$ . If you ended up with something like $\frac{\sqrt{x}}{x+1}$ , it's incorrect. You need to consider what to multiply the fraction by to eliminate the square root. Specifically, if we multiply the fraction $\frac{\sqrt{x}}{\sqrt{x}}$ , it won't help much because:

As you can see, it didn't help much. The solution can be found in the following example:

To remove the square root in the denominator, we need to multiply the entire fraction by $\frac{1-\sqrt{x}}{1-\sqrt{x}}$ .

Rationalize the fraction $\frac{x}{2*(\sqrt{x}-\sqrt{y})}$ . We will solve this example in the same way as the previous one. In the denominator, we have two terms (which we multiply by two, but that’s not important right now). These terms are √x-√y. To get rid of them, we multiply the entire fraction by $\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}$ :

When rationalizing, we must not forget about the conditions. In the last case, the variable x must not be less than zero, and also x ≠ 1.

Examples

1) Rationalize the fraction $\frac{x}{\sqrt{x}}$ :

The denominator has only one term, so we just need to multiply the entire fraction by $\frac{\sqrt{x}}{\sqrt{x}} \rightarrow \frac{x}{\sqrt{x}}*\frac{\sqrt{x}}{\sqrt{x}}=\frac{x*\sqrt{x}}{x}=\sqrt{x}$

2) Rationalize the fraction $\frac{45+5x}{4*\sqrt(x+y)}$ :

In the denominator, we have a product, so we just need to multiply the entire fraction by the square root $\frac{\sqrt{x+y}}{\sqrt{x+y}} \rightarrow \frac{45+5x}{4*\sqrt{x+y}}*\frac{\sqrt{x+y}}{\sqrt{x+y}} = \frac{(45+5x)*\sqrt{x+y}}{4*(x+y)}$ .

3) Rationalize the fraction $\frac{5}{\sqrt{z}-x}$

This is not such an easy example. Multiplying the entire fraction by $\frac{\sqrt{z}}{\sqrt{z}}$ in this case won’t help. Although I have already explained this once in this article, I will remind you → we need to look carefully at the formula (a+b)*(a-b)=a^2-b^2 to find the solution. We need to multiply the entire fraction by $\frac{\sqrt{z}+x}{\sqrt{z}+x}$ . If we do this, we get the result: $\frac{5}{\sqrt{z}-x}*\frac{\sqrt{z}+x}{\sqrt{z}+x}=\frac{5*(\sqrt{z}+x)}{z-x^2}$

4) Rationalize the fraction $\frac{2*\sqrt{3}+\sqrt{2}}{2*\sqrt{2}-\sqrt{3}}$ :

Higher Roots

So far, we have explained how to rationalize fractions where there is a square root. But of course, there are also fractions with higher roots in the denominator. The process of rationalizing such fractions is almost the same.

1) Rationalize the fraction $\frac{1}{\sqrt[3]{x}}$

If there were not the third root but just a normal square root, it would be enough to multiply the fraction by $\frac{\sqrt{x}}{\sqrt{x}}$ , but this won't work in this case. We need to multiply the entire fraction by $\frac{\sqrt[3]{x^2}}{\sqrt[3]{x^2}}$ . The entire example would then look like this:

2) Rationalize the fraction $\frac{5+y}{4\sqrt[3]{y+x}}$

Higher roots can also be adjusted, of course. Let's try to rationalize the fraction $\frac{4+y}{4\sqrt[n]{4-y}}$

This should be enough as an introduction to rationalizing fractions. If you encounter any uncertainties, feel free to ask in the comments.