Simplifying square-root expressions | Mathematics I | High School Math | Khan Academy

Let's get some practice simplifying radical expressions that involve variables. So let's say I have ( 2 \times \sqrt{7x} \times 3 \times \sqrt{14x^2} ). Pause the video and see if you can simplify, taking any perfect squares out, multiplying, and then taking any perfect squares out of the radical sign.

Well, let's first just multiply this thing so we can change the order of multiplication. This is going to be the same thing as ( 2 \times 3 \times \sqrt{7x} \times \sqrt{14x^2} ). So this is going to be equal to ( 6 \times ) and then the product of two radicals can be viewed as the square root of the product. So, ( 6 \times \sqrt{7x \times 14x^2} ).

Actually, let me factor 14. 14 is ( 2 \times 7 \times x^2 ). Let me extend my radical sign a little bit. The reason why I didn't multiply it out is because we could have done that. ( x \times x^2 ) is ( x^3 ), and we could have said, "All right, ( 7 \times 14 ) is what, ( 98 )?" We could have done that, but when you're trying to factor out perfect squares, it's actually easier if it's in this factored form.

From a variable point of view, you could view this as a perfect square already. ( 14 ) is not a perfect square, ( 7 ) isn't a perfect square, but ( 7 \times 7 ) is ( 49 ). Let's rewrite this a little bit to see what we can do. This is going to be ( 6 \times \sqrt{49 \times x^2} \times \sqrt{2x} ).

Now, we could take the square root of the perfect squares. This comes straight out of our exponent properties, but what's valuable about this is we now see this as ( 6 \times 7x \times \sqrt{2x} ). The key thing to appreciate is that the radical of products is the same thing as the product of the square roots.

Even in this step that I did here, you could say that ( \sqrt{49x^2} = \sqrt{49} \times \sqrt{x^2} = 7 \times x ). Let's do another one of these.

So let's say I have ( \sqrt{2a} \times \sqrt{14a^3} \times \sqrt{5a} ). Like always, pause this video and see if you can simplify this on your own. Multiply them and then take all the perfect squares out of the radical.

So let’s multiply first. This is going to be the same thing as ( \sqrt{2 \times 14 \times 5} ). Let me factor it. 14 can be written as ( 2 \times 7 ).

So we have ( 2 \times (2 \times 7) \times 5 \times a \times a^3 \times a = \sqrt{(2 \times 2) \times (a^4)} \times \sqrt{(35a)} ). Now, the principal root of 4 is 2, the principal root of ( a^4 ) is ( a^2 ), and we're going to have that times ( \sqrt{35a} ).

Now, let's do one more example, and this time we're going to involve two variables, which as you’ll see, isn’t that much more complicated.

So let's simplify ( \sqrt{72x^3z^3} ). The key is can we factor? 72 is not a perfect square, but if you factor it, you get ( 36 \times 2 ).

36 is a perfect square, and likewise, ( x^3 ) and ( z^3 ) are not perfect squares, but they each have an ( x^2 ) and ( z^2 ) in them. So let me rewrite this. This is the same thing as ( \sqrt{36 \times x^2 \times z^2} \times \sqrt{(2 \times 2 \times x \times x \times z)} ).

2 is left, ( x^3/x^2 = x ), ( z^3/z^2 = z ). So this is ( \sqrt{36 \times x^2 \times z^2} ) giving us ( 6xz \sqrt{2xz} ).

And we are done!