McGraw Hill Glencoe Algebra 1, 2012
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McGraw Hill Glencoe Algebra 1, 2012 View details
4. Solving Quadratic Equations by Completing the Square
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Exercise 9 Page 576

Substitute the information given into the rectangle area formula, then simplify.

Length: 18ft.
Width: 8ft.

Practice makes perfect

When we are given the area of a rectangle, we generally need to use the formula for the area of a rectangle. In this case we need to write expressions for the length, l, and width, w.

Verbal Expression Algebraic Expression
the length is 10 ft more than the width l = 10 + w

Now that we have an algebraic expression, we can substitute. Then, let's see where we can go from there.

A = l * w
144 = ( 10+w)w
â–¼
Simplify
144 = 10w+w^2
0=10w+w^2-144
0 = w^2 + 10w-144
w^2 + 10w-144=0

Now, we have a quadratic equation that is equal to zero. Let's look for two factors of -144 that have a sum of 10. Since ac is negative, the sign of the factors will be different. Furthermore, since b is positive, the factor with the larger magnitude will be positive too.

Factors Sum of factors
-1, 144 143
-2, 72 70
-3, 48 45
-4, 36 32
-6, 24 18
-8, 18 10

We have found our factors that have a sum of 10. Since our leading coefficient is 1, we can use -8 and 18 directly as part of the factoring. w^2 + 10w-144=(w-8)(w+18) Let's set each factor equal to zero and solve for w.

(w-8)(w+18)=0
â–¼
Solve for w
lcw-8=0 & (I) w+18=0 & (II)
lw=8 w+18=0
lw=8 w=-18

Since w represents a measurement of distance, we can ignore the negative solution and say the width of the rectangle is 8 ft. To find the length, we can add 10 to the width and get 18ft.