McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
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Exercise 55 Page 441

For any real number a contained in a radical such that sqrt(a^n), the root is a if n is odd and |a| if n is even.

m^2sqrt(6mp)/p^6

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(6m^5)/sqrt(p^(11))=sqrt(6m^5/p^(11)) 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(6m^5)/sqrt(p^(11))
sqrt(6m^5/p^(11))

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

sqrt(6m^5/p^(11))
sqrt(6* m^4* m/p^(10)* p)
sqrt(6* m^(2* 2)* m/p^(5* 2)* p)
sqrt(6*(m^2)^2* m/(p^5)^2* p)
sqrt((m^2)^2 * 6* m/(p^5)^2* p)

Let's stop here for a moment and consider the fact that we need to have a rationalized denominator. If we simplified all of the perfect squares as-is, we would be left with sqrt(p) in the denominator. To avoid this, we can multiply the numerator and denominator by a factor that will create a perfect square, which in this case is p.

sqrt((m^2)^2 * 6* m/(p^5)^2* p)
sqrt((m^2)^2 * 6* m* p/(p^5)^2* p * p)
sqrt((m^2)^2 * 6* m* p/(p^5)^2* p^2)
â–¼
Simplify
sqrt((m^2)^2/(p^5)^2* p^2)*sqrt(6* m* p)
m^2/p^6sqrt(6* m* p)
m^2sqrt(6* m* p)/p^6
m^2sqrt(6mp)/p^6