CPM Homework Help : AC Problem 8-109

Solve the following problem using any method. Write your solution as a sentence.

The length of a rectangle is $5$ cm longer than twice the length of the width. If the area of the rectangle is $403$ square centimeters, what is the width?

Step 1:

Make a diagram and label the sides.

Step 2:

Write an equation to solve for the width ( $x$ ).

Step 3:

Distribute and solve for $x$ .

Step 4:

Use the quadratic formula to solve for $x$ .

Partial Answer:

$x=\frac{52}{4}=13 \;\;\;\;\;\;\;\;\; x=\frac{-62}{4}=-15.5$

Can't have a negative length.

left edge $x$

rectangle interior $403 \text{ cm}^2$

bottom edge $2x+5 \text{ cm}$

$(2x+5)(x)=403$

$2x^2+5x=403$
$2x^2+5x−403=0$

$x=\frac{-5\pm\sqrt{5^{2}-4(2)(-403)}}{2(2)}$

$x=\frac{-5+57}{4} \; \text{and} \; x=\frac{-5-57}{4}$