Respuesta :
Let us call L the length of the hall and w its width.
Then the area of the hall is
[tex]L\times w[/tex]and we are told that is equal to 896 ft; therefore,
[tex]L\times w=896[/tex]Now at this point, we have to remember that we are told that the width is 4 less than length; therefore, w = L -4 and the above equation becomes
[tex]L\times(L-4)=896[/tex]Now we have to solve for L.
Expanding the left- hand side of the above equation gives
[tex]L^2-4L=896[/tex]subtract 896 from both sides to get
[tex]L^2-4L-896=0[/tex]Now the quadratic formula says that if we have an equation of the form
[tex]ax^2+bx+c=0[/tex]then
[tex]L=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Now in our case we have L instead of L and b = -4, c = -896; therefore,
[tex]L=\frac{-(-4)\pm\sqrt[]{(-4)^2-4(1)(-896)}}{2(1)}[/tex]Simplifying the right - hand side of the above equation gives
[tex]L=\frac{4\pm\sqrt[]{16+3584}}{2}[/tex][tex]x=\frac{4}{2}\pm\frac{\sqrt[]{16+3584}}{2}[/tex]which gives
[tex]L=2\pm30[/tex]Therefore the two values of L we get are
[tex]L=2+30=32[/tex][tex]L=2-30=-28[/tex]SInce lengths cannot be negative, the value L = 32 is the right answer.
Hence, the length of the hall is 32 ft.
