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:
[tex]area = length \times width[/tex]
Substituting the given values, we get
[tex]48 = (2+x) \times x\\\\48 = 2x + x^2\\\\x^2+2x - 48 = 0[/tex]
The above equation is used to find the width "x"
Let us solve the above equation by quadratic formula
[tex]\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
[tex]For\ a=1,\:b=2,\:c=-48\\\\x=\frac{-2\pm \sqrt{2^2-4\cdot \:1\left(-48\right)}}{2\cdot \:1}[/tex]
[tex]x = \frac{-2 \pm \sqrt{4+192}}{2}\\\\x = \frac{-2 \pm \sqrt{196}}{2}\\\\x = \frac{-2 \pm 14}{2}[/tex]
Thus the two values of "x" are:
[tex]x = \frac{-2 + 14}{2} \text{ or } x = \frac{-2 - 14}{2}\\\\x = \frac{12}{2} \text{ or } x = \frac{-16}{2}\\\\x = 6 \text{ or } x = -8[/tex]
Since width cannot be negative, x = -8 is not a solution
Therefore, x = 6
Thus the width of rectangle is 6
