A picture frame holds a 4-in. by 6-in. photograph. the frame adds a border x inches wide around the three sides of the photo. on the fourth side the frame forms a border that is 3x - 0.5 in. wide. the combined area of the photograph and the frame is 80.5 in^2. write a quadratic equation for the combined area. Then use the quadratic equation formula to find x

1. The quadratic equation is 3x² +17.5x - 58.5 = 0

2. The values of x are 2.38 and -8.21, but we only take 2.38 for our problem.

3. The dimensions of the combined area are approximately 6.38 inches and 12.64 inches.

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Photograph dimensions = 4 inches * 6 inches

Frame dimensions = x inches * 3x - 0.5 inches

Area of the combined area of the photograph and the frame = 80.5 inches²

2. Write a quadratic equation for the combined area. Then use the quadratic equation formula to find x.

One side of the combined area of the photograph and the frame = 4 + x inches

Second side of the combined area of the photograph and the frame = 6 + 3x - 0.5 inches

Now, we can write the quadratic equation this way:

Combined area of the photograph and the frame = One side * Second side, replacing with the values we know:

80.5 = (4 + x) * (6 + 3x - 0.5)

80.5 =  (4 + x) * (5.5 + 3x)

80.5 = 22 + 5.5x + 12x + 3x²

3x² +17.5x - 58.5 = 0

Using the quadratic formula, where, a = 3, b = 17.5 and c = - 58.5, we have:

x = (-b +/- √b² - 4ac)/2a

x = (-17.5 +/- √17.5² - 4 * 3 * -58.5) 2 * 3

x = (-17.5 +/- √306.25 + 702)/6

x = (-17.5 +/- √1,008.25)/6

x = (-17.5 +/- 31.75)/6

x₁ = 14.25/6 = 2.38

x₂ = -49.25/6 = -8.21

We took the value of x₁ because x₂ is negative and we can't have a negative value for the length or the width, therefore:

80.5 = (4 + x) * (5.5 + 3x)

Replacing x, we have:

80.5 = (4 + 2.38) * (5.5 + 3 * 2.38)

80.5 = 6.38 * 12.64

80.5 ≅ 6.38 * 12.64

step-by-step explanation:

none of the above.

for one of the equations to be used in the process, there has to be a term in the equation since squared is .

since there is only a term in the first equation, we cannot derive a from completing the square.

solving with completing the square