Square Root Product Property - Definition & Examples

(Video) Product Property of Square Roots

by Learn Math Tutorials

Here's a video by Learn Math Tutorials explaining the product property of rational square roots and showing examples.

$\large{\text{Summary}}$

The product property of square roots states that the product of square roots is equal to the square root of the product. $$\sqrt{a}\times\sqrt{b}=\sqrt{a\times b}$$

Let's show this with $9$ and $16$. $$\sqrt9\times\sqrt{16}=3\times4=12\\ \sqrt9\times\sqrt{16}=\sqrt{9\times16}=\sqrt{144}=12$$

We can use this property to help us simplify square roots, by pulling out factors that are perfect square roots. Take $\sqrt{27}$ for example. $$\sqrt{27}=\sqrt{9\times3}$$ Because of the product property, we can write this as the product of separate square roots. $$\sqrt{9\times3}=\sqrt9\times\sqrt3=3\sqrt3$$ So we were able to simplify $\sqrt{27}=3\sqrt{3}$.

Product Property of Square Roots

The Product Property of Square Roots states that the square root of a product is equal to the product of the square roots of each of the factors.

Think of the Multiplication Property of of Radicals.

Here's another way to explain it, using a product with two factors:

For example: $$\begin{align} \sqrt{10}&= \sqrt{2 \cdot 5} = \sqrt{2} \cdot \sqrt{5}\\ \sqrt{42} &= \sqrt{6 \cdot 7} = \sqrt{6} \cdot \sqrt{7} = \sqrt{2 \cdot 3} \cdot \sqrt{7} = \sqrt{2} \cdot \sqrt{3} \cdot \sqrt{7}\\ \sqrt{28} &= \sqrt{4 \cdot 7} = \sqrt{4} \cdot \sqrt{7} = 2\sqrt{7} \end{align}$$

$($It is important to note that you cannot split up radicals with sums and differences in the same way. In general, $\sqrt{x\pm y} \neq \sqrt{x}\pm \sqrt{y}.)$

So once we have some square root, say $\sqrt{x}$, we want to use this number like we use all other numbers. One thing which we know how to do is just multiplying them! So the natural question arises, how do we multiply $\sqrt{x}$ and $\sqrt{y}$? We know $\sqrt{x} \times \sqrt{x} = x$, but what if $x$ and $y$ are different?

The product rule is simply $\sqrt{x} \sqrt{y} = \sqrt{xy}$.

We can see this with a relatively simple case. Consider $y = 4$. Notice that $\sqrt{y} = 2$. So, we want to show that $\sqrt{x}\sqrt{4} = 2 \sqrt{x} = \sqrt{4x}$.

We can see this with a small picture. So consider a square of area $x$. Then, the sides of the square will have length $\sqrt{x}$.

Now, let us extend this picture. Suppose we make a bigger square with side lengths of $2\sqrt{x}$. We can notice that when we double the side lengths, we can fit four of the squares from above inside our new square. The two squares in the picture are the same size, but the one on the left is just showing that we can fit four squares of area $x$ inside our bigger square (since the side lengths are doubled).

Now, we know the area of this square is $x+x+x+x = 4x$, so we can say the side length of the square is $\sqrt{4x}$. We already know, however, that the side length is $2 \times \sqrt{x}$, so we have shown that $\sqrt{4}\times \sqrt{x} = 2 \times \sqrt{x} = \sqrt{4x}$.

This reasoning can work with any radicals, but is harder to visualize. The takeaway is that $\sqrt{x} \sqrt{y} = \sqrt{xy}$. We can try practicing with a question

The product rule is not only useful for multiplying square roots, but also for simplifying them! Let's take $\sqrt{8}$ as an example. Then $$\sqrt{8} = \sqrt{2 \times 4} = \sqrt{2} \times \sqrt{4} = 2 \times \sqrt{2}.$$ This helps us take out the $2$ from inside the square root, and the resulting number is often a bit easier to work with. So the intuition is to try and remove as many perfect squares which are inside the square root (in this case, we had a $4$ on the inside). Later on, you will get to a concept called factoring, which is very similar to this property with square roots. But for now, let's just stick with these square roots.

Example: Product Property $\\$

Say you're given a square root:

You can "split" the square root between each element of the product.

Here is an example with a variable: $$\sqrt{10x} = \sqrt{10}\times\sqrt{x}$$

Remember the product rule for raising a number to a power? This is the same application as that. Recall that taking the square root of a number is the same as taking that number to the power of $\frac{1}{2}$. To see how: $$\sqrt{4\times 16} = (4\times 16)^{\frac{1}{2}}=4^{\frac{1}{2}}\times 16^{\frac{1}{2}}=\sqrt{4}\times \sqrt{16}$$