Simplifying radicals containing variables

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. 

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


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

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

As you can see here, the answer is always x

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

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

$$ \sqrt[3]{-125x^{12}y^{15}} $$

 Try  to  write  the  expression  inside  the  radical as 

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

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


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

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

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



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