mathstudent55

You need to go back to part a. Part a. asks for an equation. You wrote the correct equation in part b. In part b., you are asked to solve the equation.

First, you need to complete part b. Your equation is correct, but now you need to solve the quadratic equation for x.

Part b.

x^2 - 4x = 165

x^2 - 4x - 165 = 0

(x - 15)(x + 11) = 0

x - 15 = 0 or x + 11 = 0

x = 15 or x = -11

The quadratic equation has two solutions, 15 and -11.

Part c.

The solutions are 15 and -11. The length of the side of a rectangle cannot be a negative number, so the solution x = -11 from part b. must be discarded. We only use the solution x = 15. The problem told you to use x for the length, so x = 15 means the length is 15 cm. The width is x - 4, so substituting 4 for x, we get x - 4 = 15 - 4 = 11. The width is 11 cm.

Answer: The length is 15 cm, and the with is 11 cm. The solution x = -11 must be discarded since the length of a rectangle cannot be a negative number.