Question

the painting shown at the right has an area of \\(300\text{ in}^2\\). what is the value of \\(x\\)?

\\(x = \square\text{ in.}\\)
(simplify your answer. round the final answer to the nearest hundredth as needed. round all intermediate values to the nearest hundredth as needed.)

Explanation:

Set up the area equation

We are given a rectangular painting with:

• Width: \(w = 2x + 7\) inches
• Height: \(h = x\) inches
• Area: \(A = 300\text{ in}^2\)

Using the Quadratic Word Problems concept, we set up the area equation:

$$x(2x + 7) = 300$$

Write in standard quadratic form

Expand and rearrange the equation into standard form \(ax^2 + bx + c = 0\):

$$2x^2 + 7x - 300 = 0$$

Solve the quadratic equation

We apply the quadratic formula to solve for \(x\):

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

Substitute \(a = 2\), \(b = 7\), and \(c = -300\):

$$x = \frac{-7 \pm \sqrt{7^2 - 4(2)(-300)}}{2(2)}$$

$$x = \frac{-7 \pm \sqrt{49 + 2400}}{4}$$

$$x = \frac{-7 \pm \sqrt{2449}}{4}$$

Calculate the numerical values

Calculate the square root of \(2449\):

$$\sqrt{2449} \approx 49.487$$

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

$$x_1 = \frac{-7 + 49.487}{4} = \frac{42.487}{4} \approx 10.62$$

$$x_2 = \frac{-7 - 49.487}{4} = \frac{-56.487}{4} \approx -14.12$$

Determine the valid dimension

Since a physical length must be positive, we discard the negative solution.

Rounding to the nearest hundredth, we get:

$$x \approx 10.62$$

Answer:

The painting shown at the right has an area of \(300\text{ in}^2\). What is the value of x?

\(x =\) <blank>\(10.62\)</blank> in.