Exercise 28 Page 571

We have a rectangular deck with an area of 250 square feet. We want to find the length and width of the deck. Recall the area of a rectangle is length times width. A= l w ⇒ 250= l w We know that the length should be 5 feet longer than twice the width. Let's call the width x. Then, the length is 2x+5. Let's rewrite our area formula using these values. 250=( 2x+5)( x) ⇒ 250=( 2x+5)( x)We now have an equation with one variable. If we can solve for the width x, then we can find the length. Let's start by putting our equation in standard form.

250=(2x+5)(x)

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Simplify

Distr

Distribute x

250=2x^2+5x

SubEqn

LHS-250=RHS-250

0=2x^2+5x-250

Now, we can factor the right hand side of the equation so that we can apply the Zero-Product Property.

0=2x^2+5x-250

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Factor

WriteSum

Write as a sum

0=2x^2+25x-20x-250

FactorOut

Factor out x

0=x(2x+25)-20x-250

SplitIntoFactors

Split into factors

0=x(2x+25)-20x-(10)(25)

FactorOut

Factor out - 10

0=x(2x+25)-10(2x+25)

FactorOut

Factor out 2x+25

0=(x-10)(2x+25)

Now that the equation is in factored form, let's apply the Zero-Product Property.

0=(x-10)(2x+25)

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Solve for x

ZeroProdProp

Use the Zero Product Property

0=x-10 & (I) 0=2x+25 & (II)

AddEqn

(I): LHS+10=RHS+10

10=x & (I) 0=2x+25 & (II)

SubEqn

(II): LHS-25=RHS-25

10=x & (I) -25=2x & (II)

DivEqn

(II): .LHS /2.=.RHS /2.

10=x & (I) - 252=x & (II)

(I), (II): Rearrange equation

x=10 & (I) x=- 252 & (II)

We can see that x=- 252 and x=10. However, recall that x represents width. Since it does not make sense to have a negative value as a width, x=10 is the only correct value. Now that we have width, we can solve for length.

l = 2x+5

Substitute

x= 10

l = 2( 10)+5

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Simplify right-hand side

Multiply

Multiply

l = 20+5

AddTerms

Add terms

l = 25

We have found that the width of the rectangle is 10 feet and the length is 25 feet.