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