Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
Mid-Chapter Quiz
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Exercise 33 Page 632

Use the fact that conjugates are the sum and difference of the same two terms.

See solution.

Practice makes perfect
Let's start by considering a few examples of fractions with radical expressions in their denominators. 2/sqrt(3), 18/9-sqrt(5), 8/sqrt(7)+sqrt(2) The first expression has only one term in the denominator. Therefore, we can rationalize its denominator by multiplying this expression by sqrt(3)sqrt(3). 2/sqrt(3)* sqrt(3)/sqrt(3)=2sqrt(3)/3

However, in the denominators of the other two expressions, there are two terms. To rationalize these denominators, we need to multiply each expression by a fraction whose numerator and denominator are conjugates to the denominators of the corresponding expressions. Recall that conjugates are the sum and difference of the same two terms. Conjugates: a+ b and a- b Using this information, we can find conjugates to the denominators of each expression.

Fraction 18/9-sqrt(5) 8/sqrt(7)+sqrt(2)
Denominator 9- sqrt(5) sqrt(7)+ sqrt(2)
Conjugate 9+ sqrt(5) sqrt(7)- sqrt(2)

The product of conjugates is a difference of squares of the terms. Product: (a+b)(a-b)= a^2- b^2 Let's multiply each expression by the corresponding fraction and simplify.

Multiplication by the Expression 18/9-sqrt(5)* 9+sqrt(5)/9+sqrt(5) 8/sqrt(7)+sqrt(2)* sqrt(7)-sqrt(2)/sqrt(7)-sqrt(2)
Product of Conjugates 18(9+sqrt(5))/9^2-(sqrt(5))^2 8(sqrt(7)-sqrt(2))/(sqrt(7))^2-(sqrt(2))^2
Simplification 18(9+sqrt(5))/76 8(sqrt(7)-sqrt(2))/47

As we can see, the denominators are now rational values, so we reached our goal. We can conclude that conjugates are used to rationalize denominators that have two terms. When multiplying conjugates, we get the difference of two squares which eliminates the radicals.