McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
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Exercise 55 Page 83

Use the area formula with A=154 and l=w+3. Do not forget to set it equal to zero!

14 in.

Practice makes perfect
We know that the painting is 3 inches longer than it is wide. This means that if we denote the width as w, we can write the length as l = w+3. We also know that the area of the painting is 154 square inches. Let's substitute our information into the equation for the area of a rectangle and solve for w.
A = l * w
154 = ( w+3)w
154 = w^2 + 3w
0 = w^2 + 3w - 154
In order to solve, we need to factor the expression and use the Zero Product Property. Let's look for factors of -154 that have a sum of 3. Since -154 is negative, we need one positive and one negative factor. Furthermore, since 3 is positive, the factor with the larger absolute value is positive.
Factors Sum
-1, 154 153
-2, 77 75
-7, 22 15
-11, 14 3
Since our leading coefficient is 1, we can use -11 and 14 as our factors. w^2 + 3w - 154 = (w-11)(w+14) Now we can use the Zero Product Property to solve for w.
(w-11)(w+14) =0
lcw-11=0 & (I) w+14=0 & (II)
lw=11 w+14=0
lw=11 w=-14

In this case, w represents a unit of length, so the negative answer can be ignored. We can say the width of the painting is 11 inches. To find the length we need to add 3 to the width, so the length of the painting is 11+3 = 14 inches.