Special care must be taken when simplifying radicals containing variables. We can start with perhaps the simplest of examples.

$$\sqrt{x^2} = ???$$

Let us look at a few examples in this form. If x = 3 or x = 5, then we have

$$\sqrt{3^2} = \sqrt{9} = 3 = x$$
$$\sqrt{5^2} = \sqrt{25} = 5 = x$$
$$These \ examples\, show \ that\ \sqrt{x^2} = x$$

Now, let us look at an example where x is a negative number. Let x  = -6

$$\sqrt{(-6)^2} = \sqrt{36} = 6$$

When x is negative, the answer is not just x or -6 as we saw before. The answer is positive. To make sure that the answer is always positive, we need to take the absolute value.

$$For\ any \ number \ y,\ \sqrt{y^2} = |y|$$

Now what about the cube root of x? The cube root will behave a little differently.

$$\sqrt{x^3} = ???$$

If x  = 2 or x = -2, the answer is not always positive.

$$\sqrt{2^3} = \sqrt{8} = 2 = x$$
$$\sqrt{(-2)^3} = \sqrt{-8} = -2 = x$$

As you can see here, the answer is always x

$$\sqrt{(x)^3} = x$$

## Interesting or challenging examples of simplifying radicals containing variables

$$\sqrt{64y^{16}}$$

Try  to  write  the  expression  inside  the  radical as

$$\sqrt{(something)^{2}}$$
$$Then, \sqrt{(something)^{2}} = something$$

We will need to use some properties of exponents to do this.

$$\sqrt{64y^{16}} = \sqrt{8^2 \times (y^{8})^2} = \sqrt{[8y^{8}]^2}$$

Notice that something is equal to 8y8

$$Therefore, \sqrt{64y^{16}} = 8y^8$$

let us now conclude this lesson with the last example below

$$\sqrt{-125x^{12}y^{15}}$$

Try  to  write  the  expression  inside  the  radical as

$$\sqrt{(something)^{3}}$$

$$Then, \sqrt{(something)^{3}} = something$$

$$\sqrt{-125x^{12}y^{15}} = \sqrt{(-5)^3(x^4)^3(y^5)^3}$$

$$\sqrt{-125x^{12}y^{15}} = \sqrt{[(-5)(x^4)(y^5)]^3}$$

$$Therefore, \sqrt{-125x^{12}y^{15}} = -5x^4y^5$$

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