Multiplying and Dividing Square Roots, Rationalizing the Denominator

Square Roots
Multiplication and Division

At some point square roots should no longer be considered an operation but rather the most efficient way to express a number. For example, the best way to write one hundred trillion is $1\times {10}^{14}$. The best way to express the number times itself that is two is as $\sqrt{2}.$

That provides insight when we consider multiplying a rational number and an irrational number together. It is not confusing for some irrational numbers, like π. Nobody confused 3π because we understand that symbol is the best way to write the number. There’s not a way to rewrite multiples of π other than by writing the multiple in front.

However, $3\sqrt{2}$ is often written as $\sqrt{6}$. There are reasons explained by the order of operations which tell us why this is false, but understanding what the square root of two is perhaps offers the simplest insight.

$\sqrt{2}\approx 1.414$

$3\sqrt{2}=\sqrt{2}+\sqrt{2}+\sqrt{2}$

$3\sqrt{2}\approx 1.414+1.414+1.414$

4.242

The square root of six is approximately 2.449. Not the same thing at all.

 

The following, however, is true:

$\sqrt{2}\times \sqrt{3}=\sqrt{6}$

and

$\sqrt{2\times 3}=\sqrt{6}$.

The following generalization can be used. Sometimes it is best to write things one way versus another, and it is up to you to decide if rewriting an expression offers insight.

$\sqrt{ab}=\sqrt{a}\cdot \sqrt{b}$

If two numbers are both square roots you can multiply their radicands together. But you cannot multiply the radicand of a square root with rational number like we saw above.

Division is a little more nuanced, but only when your denominator is a fraction.

This generalization is true for division:

$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$

For example:

$\frac{\sqrt{8}}{\sqrt{4}}=\sqrt{2}.$

This can be calculated two ways.

$\frac{\sqrt{8}}{\sqrt{4}}=\sqrt{\frac{8}{4}}=\sqrt{2}.$

or

$\frac{\sqrt{8}}{\sqrt{4}}=\frac{2\sqrt{2}}{2}=\sqrt{2}.$

But you cannot divide rational numbers into the radicand, or the radicand of a square root into a rational number. Remember, square roots, when simplified, are the most efficient way of writing irrational numbers. If we used k to represent the square root of two, these types of confusing things would not be happening.

Nobody would confuse what is happening with $\frac{6}{k}$. We simply cannot evaluate that because 6 and k do not have common factors. When k is written as the square root of two, sometimes people just see a 2 and reduce.

The only issue with division of square roots occurs if you end up with a square root in the denominator.

$\frac{5}{\sqrt{2}}$

Denominators must be rational and the square root of two is irrational. However, there’s an easy fix. Remember that $\sqrt{2}\times \sqrt{2}=\sqrt{4},$ and $\sqrt{4}=2.$ It is also true that:

$\frac{\sqrt{2}}{\sqrt{2}}=1$ .

To Rationalize the Denominator, which means make the denominator a rational number, we just multiply as follows:

$\frac{5}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{5\sqrt{2}}{2}$

Sometimes we end up with something like this:

$\frac{5}{3\sqrt{2}}$

Three is a rational number and is perfectly okay in the denominator. If you multiply by the fraction $\frac{3\sqrt{2}}{3\sqrt{2}},$ you can still get the simplified equivalent, but you’ll have extra reducing to do at the end. Instead, just multiply by the irrational portion.

$\frac{5}{3\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{5\sqrt{2}}{6}$.

In summary, to divide or multiply with square roots, you can multiply or divide the radicands. However, if you’re multiplying or dividing rational numbers and square roots, you cannot combine the radicands and the rational numbers.

 

 

 

 

 

 

 

 

Practice Problems:

 

Perform the indicated operations:

$\begin{array}{l}1.\left(5\sqrt{7}\right)\left(3\sqrt{14}\right)\\ \\ \\ 2.\left(\sqrt{15}\right)\left(\sqrt{3}\right)\\ \\ \\ 3.\frac{3\sqrt{2}}{\sqrt{8}}\\ \\ \\ 4.\frac{\sqrt{5}}{\sqrt{3}}\\ \\ \\ \\ 5.\frac{3}{\sqrt{8}}\cdot \frac{\sqrt{2}}{6}\end{array}$