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Pearson Algebra 1 Common Core, 2011 View details

1. The Pythagorean Theorem

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Exercise 44 Page 618

Start by writing an equation for the area of the rectangular field and calculating the area of the garden.

20ft

Practice makes perfect

We are told that Joe plants a rectangular garden in the corner of his field, and the area of the garden is 60 % of the area of the field.

In order to find the longest side of the rectangular field, we will first write an equation representing the area of the field A_F.

Let's substitute l= 16+2x and w= 12+2x into the formula for the area of a rectangle, and then simplify it.

A_F=l w

SubstituteII

l= 16+2x, w= 12+2x

A_F=( 16+2x)( 12+2x)

▼

Simplify right-hand side

Distr

Distribute (16+2x)

A_F=12(16+2x)+2x(16+2x)

Distr

Distribute 12

A_F=192+24x+2x(16+2x)

Distr

Distribute 2x

A_F=192+24x+32x+4x^2

AddTerms

Add terms

A_F=192+56x+4x^2

CommutativePropAdd

Commutative Property of Addition

A_F=4x^2+56x+192

Now, let's find the area of the rectangular garden A_G in the same way.

A_G=l w

SubstituteII

l= 16, w= 12

A_G=( 16)( 12)

Multiply

A_G=192

We know that the area of the garden A_G is 60 % of the area of the field A_F. Using the fact that the fraction form of 60 % is 60100, we can write an equation to represent the relation between A_G and A_F. A_G=60/100A_F We will now substitute A_F= 4x^2+56x+192 and A_G= 192 into this equation, and simplify it as much as possible.

A_G=60/100A_F

SubstituteII

A_F= 4x^2+56x+192, A_G= 192

192=60/100( 4x^2+56x+192)

▼

Simplify

MultEqn

LHS * 100=RHS* 100

19200=60(4x^2+56x+192)

DivEqn

.LHS /60.=.RHS /60.

320=4x^2+56x+192

SubEqn

LHS-320=RHS-320

0=4x^2+56x-128

RearrangeEqn

Rearrange equation

4x^2+56x-128=0

FactorOut

Factor out 4

4(x^2+14x-32)=0

DivEqn

.LHS /4.=.RHS /4.

x^2+14x-32=0

Note that we have a quadratic equation. We will now factor the left-hand side.

x^2+14x-32=0

WriteDiff

Write as a difference

x^2+16x-2x-32=0

FactorOut

Factor out x

x(x+16)-2x-32=0

FactorOut

Factor out -2

x(x+16)-2(x+16)=0

FactorOut

Factor out x+16

(x+16)(x-2)=0

Finally, we will apply the Zero-Product Property.

(x+16)(x-2)=0

ZeroProdProp

Use the Zero Product Property

lcx+16=0 & (I) x-2=0 & (II)

SubEqn

(I): LHS-16=RHS-16

lx=-16 x-2=0

AddEqn

(II): LHS+2=RHS+2

lx=-16 x=2

The solutions to the equation are x=-16 and x=2. However, since x represents a length, it should be positive. Therefore, we can find the length of the longest side of the field when x= 2 by substituting 2 into the expression for this side length, 16+2x.

16+2x

Substitute

x= 2

16+2( 2)

Multiply

16+4

AddTerms

Add terms

The longest side of the field is 20 feet.