Question

A rectangle is 2 feet longer than it is wide. The area of the rectangle is 48 square feet. Write and solve an equation that can be used to find the width of the rectangle

Answer 1

The width of rectangle is 6

Solution:

Let the width of rectangle be "x"

A rectangle is 2 feet longer than it is wide

Therefore, length = 2 + width

length = 2 + x

The area of rectangle is 48 square feet

The area of rectangle is given by formula:

`area = length * width`

Substituting the given values, we get

`48 = (2+x) * x\\\\48 = 2x + x^2\\\\x^2+2x - 48 = 0`

The above equation is used to find the width "x"

Let us solve the above equation by quadratic formula

`\text {For a quadratic equation } a x^(2)+b x+c=0, \text { where } a \\eq 0\\\\x=\frac{-b \pm \sqrt{b^(2)-4 a c}}{2 a}`

`For\ a=1,\:b=2,\:c=-48\\\\x=(-2\pm √(2^2-4\cdot \:1\left(-48\right)))/(2\cdot \:1)`

`x = (-2 \pm √(4+192))/(2)\\\\x = (-2 \pm √(196))/(2)\\\\x = (-2 \pm 14)/(2)`

Thus the two values of "x" are:

`x = (-2 + 14)/(2) \text{ or } x = (-2 - 14)/(2)\\\\x = (12)/(2) \text{ or } x = (-16)/(2)\\\\x = 6 \text{ or } x = -8`

Since width cannot be negative, x = -8 is not a solution

Therefore, x = 6

Thus the width of rectangle is 6