Question

your school wants to take out an ad in the paper congratulating the basketball team on a successful season, as shown to the right. the area of the photo will be half the area of the entire ad. what is the value of x?

\\(x = \square\\)
(round to the nearest hundredth as needed. use a comma to separate answers as needed.)

Explanation:

Express the areas of the photo and the entire ad

The photo has a width of \(5\text{ in}\) and a height of \(3\text{ in}\).
Using Quadratic Word Problems, we find the area of the photo:

$$A_{\text{photo}} = 5 \times 3 = 15\text{ in}^2$$

The entire ad has a width of \(5 + x\) and a height of \(3 + x\).
The area of the entire ad is:

$$A_{\text{ad}} = (5 + x)(3 + x) = x^2 + 8x + 15$$

Set up the quadratic equation

The area of the photo is half the area of the entire ad:

$$A_{\text{photo}} = \frac{1}{2} A_{\text{ad}}$$

$$15 = \frac{1}{2}(x^2 + 8x + 15)$$

Multiply both sides by 2:

$$30 = x^2 + 8x + 15$$

Subtract 30 from both sides to write in standard form:

$$x^2 + 8x - 15 = 0$$

Solve using the quadratic formula

Using Quadratic Formula Application with \(a = 1\), \(b = 8\), and \(c = -15\):

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

$$x = \frac{-8 \pm \sqrt{8^2 - 4(1)(-15)}}{2(1)}$$

$$x = \frac{-8 \pm \sqrt{64 + 60}}{2}$$

$$x = \frac{-8 \pm \sqrt{124}}{2}$$

Calculate the numerical values

Simplify the square root:

$$\sqrt{124} \approx 11.1355$$

Calculate the two possible values for \(x\):

$$x_1 = \frac{-8 + 11.1355}{2} \approx 1.5678 \approx 1.57$$

$$x_2 = \frac{-8 - 11.1355}{2} \approx -9.57$$

Since a physical width \(x\) must be positive, we discard the negative solution.
Thus, \(x \approx 1.57\).

Answer:

\(x =\) <blank>1.57</blank>