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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

2. Simplifying Radicals

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Exercise 77 Page 625

Practice makes perfect

a

We are given the equation that gives us the radius r of a circle with area A.

r=sqrt(A/π) We need to find the radius for the given area and write the answer as a simplified radical. Let's start by substituting A= 50 square feet into the formula. Then we will use the Division Property of Square Roots.

r=sqrt(A/π)

A= 50

r=sqrt(50/π)

sqrt(a/b)=sqrt(a)/sqrt(b)

r = sqrt(50)/sqrt(π)

SplitIntoFactors

Split into factors

r= sqrt(25 * 2)/sqrt(π)

sqrt(a* b)=sqrt(a)*sqrt(b)

r=sqrt(25)* sqrt(2)/sqrt(π)

Simplify root

r=5(sqrt(2)/sqrt(π))

Since the radicand in the denominator is not a perfect square, we need to rationalize the denominator to remove the radical. To do so we will multiply the numerator and the denominator by the same radical expression.

r=5(sqrt(2)/sqrt(π))

Multiply by sqrt(π)/sqrt(π)

r=5(sqrt(2)/sqrt(π)* sqrt(π)/sqrt(π))

sqrt(a)*sqrt(b)=sqrt(a* b)

r=5(sqrt(2 π)/sqrt(π * π))

ProdToPowTwoFac

a* a=a^2

r=5(sqrt(2 π)/sqrt(π^2))

SqrtPowToNumber

sqrt(a^2)=a

r=5(sqrt(2 π)/π)

We can write the radius as a decimal.

r=5(sqrt(2 π)/π)

Use a calculator

r = 3.989422 ...

Round to 2 significant digit(s)

r ≈ 3.99

The radius is about 3.99 feet.

b

Proceeding in the same way, we will calculate the radius for the given area. In this case we will use the Multiplication Property of Square Roots.

r=sqrt(A/π)

A= 32

r=sqrt(32/π)

sqrt(a/b)=sqrt(a)/sqrt(b)

r = sqrt(32)/sqrt(π)

SplitIntoFactors

Split into factors

r= sqrt(4 * 8)/sqrt(π)

sqrt(a* b)=sqrt(a)*sqrt(b)

r=sqrt(4)* sqrt(8)/sqrt(π)

▼

Simplify

SplitIntoFactors

Split into factors

r = sqrt(4)*sqrt(4 * 2)/sqrt(π)

sqrt(a* b)=sqrt(a)*sqrt(b)

r = sqrt(4)sqrt(4)sqrt(2)/sqrt(π)

Simplify root

r= 2 * 2( sqrt(2)/sqrt(π))

Multiply

r= 4(sqrt(2)/sqrt(π))

Now, we can rationalize the denominator to remove the radical.

r= 4(sqrt(2)/sqrt(π))

Multiply by sqrt(π)/sqrt(π)

r=4(sqrt(2)/sqrt(π)* sqrt(π)/sqrt(π))

sqrt(a)*sqrt(b)=sqrt(a* b)

r=4(sqrt(2 π)/sqrt(π * π))

ProdToPowTwoFac

a* a=a^2

r=4(sqrt(2 π)/sqrt(π^2))

SqrtPowToNumber

sqrt(a^2)=a

r=4(sqrt(2 π)/π)

Finally, we will calculate the radius as a decimal.

r=4(sqrt(2 π)/π)

Use a calculator

r = 3.191538 ...

Round to 2 significant digit(s)

r ≈ 3.19

Therefore, the radius is about 3.19 inches.

c

Let's do the same for A= 10 square meters. We will start by rationalizing the denominator to remove the radical.

r=sqrt(A/π)

A= 10

r=sqrt(10/π)

Multiply by sqrt(π)/sqrt(π)

r=sqrt(10)/sqrt(π)* sqrt(π)/sqrt(π)

sqrt(a)*sqrt(b)=sqrt(a* b)

r=sqrt(10 π)/sqrt(π * π)

ProdToPowTwoFac

a* a=a^2

r=sqrt(10 π)/sqrt(π^2)

SqrtPowToNumber

sqrt(a^2)=a

r=sqrt(10 π)/π

Now, we need to write the radius as a decimal.

r=sqrt(10 π)/π

Use a calculator

r = 1.784124 ...

Round to 2 significant digit(s)

r ≈ 1.78

For this case, the radius is about 1.78 meters.

Subchapter links

Lesson Check p.623

Practice and Problem-Solving Exercises p.623-625

Standardized Test Prep p.625

Mixed Review p.625