Answer:
The length is [tex]8[/tex] feet.
Step-by-step explanation:
we know that
the area of the rectangle is equal to
[tex]A=xy[/tex]
where
x is the length side of rectangle
y is the width side of rectangle
In this problem we have
[tex]A=48\ ft^{2}[/tex]
so
[tex]48=xy[/tex] ----> equation A
[tex]y=x-2[/tex] ------> equation B
substitute equation B in equation A
[tex]48=x(x-2)[/tex]
[tex]x^{2}-2x-48=0[/tex]
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2}-2x-48=0[/tex]
so
[tex]a=1\\b=-2\\c=-48[/tex]
substitute in the formula
[tex]x=\frac{-(-2)(+/-)\sqrt{-2^{2}-4(1)(-48)}} {2(1)}[/tex]
[tex]x=\frac{2(+/-)\sqrt{4+192}} {2}[/tex]
[tex]x=\frac{2(+/-)14} {2}[/tex]
[tex]x=\frac{2+14} {2}=8\ ft[/tex] -----> the solution
[tex]x=\frac{2-14} {2}=-6[/tex]
