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Multiplying Radical Expressions

In this video, we will multiply radical expressions.

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Let’s start with simple expressions.

In other words,

Let’s try other examples:
\sqrt{5}\times\sqrt{7} would equal to \sqrt{35}
\sqrt{18}\times\sqrt{2} would equal to \sqrt{36}, which is a perfect square so it would be 6

Let’s try the same problem by using another method:
\sqrt{18}\times\sqrt{2} would equal to \sqrt{36} can also be written as \sqrt{9}\times\sqrt{2}\times\sqrt{2}
This also leads to the answer of 6

For \sqrt{15}\times\sqrt{45}, it would equal to \sqrt{675}
25 goes into 675 twenty-seven times
Since 5 is the square root of 25, \sqrt{675} can be written as 5\sqrt{27}
This can be further simplified into 15\sqrt{3}

Video-Lesson Transcript

Let’s go over how to multiply radical expressions.

First, let’s start with

\sqrt{3} \times \sqrt{3} = \sqrt{9} = 3
\sqrt{5} \times \sqrt{5} = \sqrt{25} = 5

So if you look, closely

\sqrt{7} \times \sqrt{7} = 7
\sqrt{12} \times \sqrt{12} = 12

Keep this in mind as we move forward in the lesson.

Let’s take a look at this

\sqrt{5} \times \sqrt{7} = \sqrt{35}

this cannot be simplified.

\sqrt{18} \times \sqrt{2} = \sqrt{36} = 6

But what if we reduce this first?

\sqrt{9} \times \sqrt{2} \times \sqrt{2}
3 \times 2

So you can multiply it across and get an answer.

Or reduce it first and multiply it out.

Let’s have this one

\sqrt{15} \sqrt{45}

I can multiply this out

\sqrt{15} \sqrt{45} = \sqrt{675}

Then we should break 675 down.

But it’s a pretty large number.

\sqrt{25} \sqrt{27}
5 \sqrt{27}
5 \sqrt{9} \sqrt{3}
5 \times 3 \sqrt{3}
15 \sqrt{3}

Let’s see what happens if we reduce it first.

\sqrt{15} \sqrt{45}
\sqrt{15} \sqrt{9} \sqrt{5}
\sqrt{15} \times (3) \sqrt{5}
3 \sqrt{75}
3 \sqrt{25} \sqrt{3}
3 \times 5 \sqrt{3}
15 \sqrt{3}

We came up with the same answer.

The number of steps in the two methods is pretty much the same.

But I dealt with smaller numbers using the second method.

Let’s look back at a point here.

\sqrt{15} \sqrt{45}
\sqrt{15} \sqrt{9} \sqrt{5}

At this point, we know that \sqrt{15} can’t be broken down.

But there’s a \sqrt{5} there.

And we know that 5 goes into 15.

So, we might as well break it down. So we’ll have

\sqrt{3} \sqrt{5} \sqrt{9} \sqrt{5}

Let’s just reorganize this

\sqrt{9} \sqrt{5} \sqrt{5} \sqrt{3}

And we’ll solve

3 \times 5 \sqrt{3}