Exercise 4 Page 570

We are making a rectangular table with an area of 10 square feet. The dimensions of a rectangle are its length and width. Recall that the area of a rectangle is length times width. A= l w ⇒ 10= l w We know that the length of the table is one foot shorter than twice its width. Let's call the width of the table x. The length can be written as l=2x-1. 10= l w ⇒ 10=( 2x-1) xWe 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 rewriting our equation in standard form.

10=(2x-1)x

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Simplify

Distr

Distribute x

10=2x^2-x

SubEqn

LHS-10=RHS-10

0=2x^2-x-10

RearrangeEqn

Rearrange equation

2x^2-x-10=0

Let's now factor the left-hand side of the above equation.

2x^2-x-10=0

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Factor

WriteDiff

Write as a difference

2x^2+4x-5x-10=0

FactorOut

Factor out 2x

2x(x+2)-5x-10=0

SplitIntoFactors

Split into factors

2x(x+2)-5x-5(2)=0

FactorOut

Factor out - 5

2x(x+2)-5(x+2)=0

FactorOut

Factor out (x+2)

(x+2)(2x-5)=0

Now that the equation is written in factored form, we will apply the Zero-Product Property.

(x+2)(2x-5)=0

ZeroProdProp

Use the Zero Product Property

lcx+2=0 & (I) 2x-5=0 & (II)

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Solve for x

SubEqn

(I): LHS-2=RHS-2

lx=- 2 2x-5=0

AddEqn

(II): LHS+5=RHS+5

lx=- 2 2x=5

DivEqn

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

lx=- 2 x= 52

We can see that x= 52 and x=-2 are solutions to the equation. However, recall that x represents width. Since it does not make sense to have a negative value as a width, x= 52 is the only correct value. Now that we have width, we can solve for length.

l = 2x-1

Substitute

x= 5/2

l = 2( 5/2)-1

DenomMultFracToNumber

2 * a/2= a

l = 5-1

SubTerms

Subtract terms

l = 4

We have found that the width of the rectangle is 52 feet and the length is 4 feet.