Answer:
[tex]W=(x-9)\ m[/tex]
Step-by-step explanation:
we know that
The area of rectangle is equal to
[tex]A=LW[/tex]
where
L is the length of rectangle
W is the width of rectangle
we have
[tex]A=(x^{2} -x-72)\ m^2[/tex]
[tex]L=(x+8)\ m[/tex]
[tex]W=\frac{A}{L}[/tex]
substitute
[tex]W=\frac{(x^{2} -x-72)}{(x+8)}[/tex]
Solve the quadratic equation in the numerator
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2} -x-72=0[/tex]
so
[tex]a=1\\b=-1\\c=-72[/tex]
substitute in the formula
[tex]x=\frac{-(-1)\pm\sqrt{-1^{2}-4(1)(-72)}} {2(1)}[/tex]
[tex]x=\frac{1\pm\sqrt{289}} {2}[/tex]
[tex]x=\frac{1\pm17} {2}[/tex]
[tex]x=\frac{1+17} {2}=9[/tex]
[tex]x=\frac{1-17} {2}=-8[/tex]
so
[tex]x^{2} -x-72=(x+8)(x-9)[/tex]
substitute in the expression of W
[tex]W=\frac{(x^{2} -x-72)}{(x+8)}[/tex]
[tex]W=\frac{(x+8)(x-9)}{(x+8)}[/tex]
simplify
[tex]W=(x-9)\ m[/tex]
