McGraw Hill Glencoe Algebra 2, 2012
MH
McGraw Hill Glencoe Algebra 2, 2012 View details
5. Operations with Radical Expressions
Continue to next subchapter

Exercise 24 Page 419

For any real number a, we have that sqrt(a^n)=a if n is an odd number, and that sqrt(a^n)=|a| if n is even.

sqrt(150x^2y^2)/5y

Practice makes perfect

Before dividing the given radical expressions, we need to answer two questions.

  1. Can the expressions be divided?
  2. If so, do absolute value symbols need to be added to the answer?

The rule regarding dividing radical expressions states that if sqrt(a) and sqrt(b) are real numbers and b≠0, then sqrt(a)÷sqrt(b)=sqrt(a÷ b). sqrt(6x^2)/sqrt(5y)=sqrt(6x^2/5y) Because we are assuming that both radicals are real numbers and we can see that the given expressions have the same index, we can divide them. Now, to answer the second question, consider the rule regarding absolute value symbols for any real number a. sqrt(a^n)= a if n is odd |a| if n is even Because our radicals have an odd root, negative numbers will not cause the expressions to be imaginary. Therefore, we do not need to worry about absolute value symbols.

sqrt(6x^2)/sqrt(5y)
sqrt(6x^2/5y)

Because there are not any perfect cubes under the radical, we just need to rationalize the denominator. To do that, we can multiply the numerator and denominator by a factor that will create a perfect cube. Let's start by finding the necessary exponents. Our goal is to have three of each factor.

sqrt(6x^2)/sqrt(5y)
sqrt(6x^2)/sqrt(5^1 y^1)
sqrt(6x^2)* sqrt(5^2y^2)/sqrt(5^1 y^1) *sqrt(5^2 y^2)
sqrt(6x^2 * 5^2 y^2)/sqrt(5^1 y^1 * 5^2 y^2)
sqrt(6x^2 * 5^2 y^2)/sqrt(5^3 y^3)

Now that we have found the factors, we can simplify the expression.

sqrt(6x^2 * 5^2 y^2)/sqrt(5^3 y^3)
sqrt(6x^2 * 25y^2)/sqrt(5^3 y^3)
sqrt(150x^2y^2)/sqrt(5^3 y^3)
sqrt(150x^2y^2)/5y