Pearson Geometry Common Core, 2011
PG
Pearson Geometry Common Core, 2011 View details
1. The Pythagorean Theorem and Its Converse
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Exercise 61 Page 498

Start by rewriting the given expression as a fraction. Then rationalize the denominator.

15sqrt(2)

Practice makes perfect
To simplify the given expression, we can first rewrite it as a fraction. 30 ÷ sqrt(2) ⇔ 30/sqrt(2) Now, we can rationalize the denominator of the quotient. To do that, we will multiply the numerator and denominator by a factor that will make the denominator a perfect square inside the square root. We will do this using the fact that we can multiply the radicands of radicals if they have the same index. If sqrt(a) and sqrt(b) are real numbers, then sqrt(a)* sqrt(b)= sqrt(ab). Let's start by finding the exponents necessary to create perfect squares in the denominator. Our goal is to have two of each factor.
30/sqrt(2)
30/sqrt(2^1)
30sqrt(2^1)/sqrt(2^1) *sqrt(2^1)
30sqrt(2^1)/sqrt(2^1* 2^1)
30sqrt(2^1)/sqrt(2^2)
Now that we've found the factors that will make the radicand of the denominator perfect squares only, we can begin to simplify the quotient.
30sqrt(2^1)/sqrt(2^2)
30sqrt(2^1)/2
30sqrt(2)/2
15sqrt(2)/1
15sqrt(2)