Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
4. Factoring to Solve Quadratic Equations
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Exercise 27 Page 571

Write the length of the rectangle as a function of the rectangle's width.

Length: 6 feet
Width: 4 feet

Practice makes perfect

We have a rectangular blanket with an area of 24 square feet. We want to find the length and width of the blanket. Recall the area of a rectangle is length times width. A= l w ⇒ 24= l w We know that the length should be 2 feet longer than the width. We can call the width x. This makes the length is x+2. Let's rewrite our area formula using these values. 24=( x+2)( x) ⇒ 24=( x+2)( x)We now have an equation with one variable. If we can solve for the width x we can find the length. Let's start by putting our equation in standard form.

24=(x+2)(x)
â–¼
Simplify
24=x^2+2x
0=x^2+2x-24

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

0=x^2+2x-24
â–¼
Factor
0=x^2+6x-4x-24
0=x(x+6)-4x-24
0=x(x+6)-4x-(4)(6)
0=x(x+6)-4(x+6)
0=(x+6)(x-4)

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

0=(x+6)(x-4)
â–¼
Solve for x
0=x+6 & (I) 0=x-4 & (II)
-6=x & (I) 0=x-4 & (II)
-6=x & (I) 4=x & (II)

(I), (II): Rearrange equation

x=- 6 & (I) x=4 & (II)

We can see that x=- 6 and x=4. However, recall that x represents width. Since it does not make sense to have a negative value as a width, x=4 is the only correct value. Now that we have width, we can solve for length.

l = x+2
l = 4+2
l = 6

We have found that the width of the rectangle is 4 feet and the length is 6 feet.