Question Video: Simplifying Algebraic Expressions with More than one Variable Involving Exponents and Square Roots | Nagwa Question Video: Simplifying Algebraic Expressions with More than one Variable Involving Exponents and Square Roots | Nagwa

# Question Video: Simplifying Algebraic Expressions with More than one Variable Involving Exponents and Square Roots Mathematics • Second Year of Secondary School

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Write β(25πΒ²πβΆ) in its simplest form.

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### Video Transcript

Write the square root of 25π squared π to the sixth power in its simplest form.

In order to simplify this expression, weβre going to recall that if the πth root of π, the πth root of π, and the πth root of π are well defined for positive integers π, then their product is the πth root of πππ. Weβre going to apply this property in reverse to allow us to write the original expression as the square root of 25 times the square root of π squared times the square root of π to the sixth power. Now, we know that the square root of 25 is simply equal to five. But what do we do with the square root of π squared and the square root of π to the sixth power?

Well, since weβre finding the square root, this is an πth root where π is even; itβs two. So we use the following property. If π is even and π is a real number, then the πth root of π to the πth power is equal to the absolute value of π. So we can actually rewrite the square root of π squared as the absolute value of π. To repeat this process for the square root of π to the sixth power, we need to rewrite π to the sixth power as π cubed squared, meaning that the square root of π cubed all squared is equal to the absolute value of π cubed.

We can now replace each root with the value we found. So the square root of 25 times the square root of π squared times the square root of π to the sixth power is five times the absolute value of π times the absolute value of π cubed. Then, since five is by its very nature nonnegative, we can use the rules for multiplying absolute value expressions to rewrite this as the absolute value of five ππ cubed. And so in its simplest form, the square root of 25π squared π to the sixth power is equal to the absolute value of five ππ cubed.

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