McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
5. Operations with Radical Expressions
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Exercise 27 Page 419

To simplify the expression, remember the Product Property of Radicals.

32a^5b^3sqrt(b)

Practice makes perfect

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

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

The rule regarding multiplying radical expressions states that if sqrt(a) and sqrt(b) are real numbers, then sqrt(a)*sqrt(b) = sqrt(a b). 2sqrt(32a^3b^5)* sqrt(8a^7b^2)=2sqrt(32a^3b^5* 8a^7b^2) Because we are assuming that both radicals are real numbers and we can see that the given expressions have the same index, we can multiply 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 Since both radicals are real numbers and the roots are even, the expressions underneath the radicals are positive. Otherwise, the radicals would be imaginary. With this in mind, let's consider the possible values of the variables a and b.

  • In the numerator, the index is even and the exponent of a and b are odd. For the expression to result in a real number, the product of a and b must be positive. Therefore, a and b must have the same sign — both positive or both negative.
  • In the second, the index is even so the expression contained in the radical must be positive.
    • The product of a^7 and b^2 must be positive — a^7 and b^2 must have the same sign.
    • Because the result of b^2 will always be positive, a^7 must also be positive.
  • If a^7 is positive, a is positive. Since a and b must have the same sign, a and b must both be positive.

This means that if we remove a or b from the radical, we will not need absolute value symbols.

2sqrt(32a^3b^5)* sqrt(8a^7b^2)
2sqrt(32a^3b^5* 8a^7b^2)
2sqrt(256a^3b^5 * a^7b^2)
2sqrt(256a^(10)b^7)

Next, let's simplify the radical expression by finding all of the perfect squares inside the radical.

2sqrt(256a^(10)b^7)
2sqrt(16^2a^(10)b^7)
2sqrt(16^2 * a^2* a^2 * a^2* a^2 * a^2 * b^2 * b^2 * b^2 * b)
2sqrt((16a^5b^3)^2 * b)
â–¼
Simplify
2sqrt((16a^5b^3)^2)* sqrt(b)
2* 16a^5b^3* sqrt(b)
32a^5b^3sqrt(b)