Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
2. Simplifying Radicals
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Exercise 57 Page 624

Practice makes perfect
a We are given the following expression.

sqrt(a)* sqrt(b)To verify if its value is 6sqrt(5) when a= 18 and b= 10, let's substitute the given values for a and b. Then we will simplify the expression.

sqrt(a)* sqrt(b)
sqrt(18)* sqrt(10)
sqrt(18* 10)
sqrt(180)

Next, we need to look for perfect-square factors of 180 so that we can further simplify the expression.

sqrt(180)
sqrt(36* 5)
sqrt(36)*sqrt(5)
6sqrt(5)

As we can see, the value of the expression for the given values of a and b is indeed 6sqrt(5).

b Now we want to find two other pairs of positive integers a and b that make the given equation true. Let's use the fact that 6sqrt(5) can be rewritten as sqrt(180) — as shown in Part A.

6sqrt(5)= sqrt(180) By the Multiplication Property of Square Roots, the product of two squares is equal to the square root of the product of the radicands. sqrt(a)* sqrt(b)= sqrt(ab) Therefore, we want to find values a and b that satisfy the following equation. sqrt(ab)&= sqrt(180) & ⇓ ab&=180 To find the possible values of a and b, we will split 180 into factor pairs. 180&=18* 10 &=20* 9 &=30* 6 &=36* 5 &=45* 4 &=60* 3 &=90* 2 The square root of any of these products is equal to 6sqrt(5). Hence, two possible pairs of values for a and b are as follows. la=4 b=45 and la=3 b=60 Note that the values of a and b can be reversed within any pair.