Question

a rectangle with an area of 120 in² has a length 8 inches longer than two times its width. what is the width of the rectangle? a 3 inches b 6 inches c 9 inches d 12 inches

Explanation:

Step1: Define the width variable

Let the width of the rectangle be $x$ inches. Then the length $l$ is $2x + 8$ inches.

Step2: Use the area formula

The area formula of a rectangle is $A=l\times w$. We know $A = 120$ square - inches. So, $120=(2x + 8)\times x$.

Step3: Expand the equation

Expand $(2x + 8)\times x$ to get $2x^{2}+8x$. So the equation becomes $2x^{2}+8x-120 = 0$.

Step4: Simplify the equation

Divide the entire equation $2x^{2}+8x - 120=0$ by 2, we get $x^{2}+4x-60 = 0$.

Step5: Factor the quadratic equation

Factor $x^{2}+4x - 60$ as $(x + 10)(x - 6)=0$.

Step6: Solve for x

Set each factor equal to zero: $x+10 = 0$ gives $x=-10$; $x - 6=0$ gives $x = 6$. Since the width cannot be negative, we discard $x=-10$.

Answer:

6 inches