find the dimensions of a rectangle whose width is 4 miles less than its length and whose area is 96 square miles

Answer:12 miles by 8 milesStep-by-step explanation:Let length be represented by xSince width is 4 miles less than the length, therefore, the width will be (x-4) milesArea=Length*WidthArea=x(x-4)=[tex]x^{2}-4x[/tex]Area is given hence[tex]x^{2}-4x=96[/tex]To write the above equation as a quadratic equation[tex]x^{2}-4x-96=0[/tex]By factorization, we find two numbers whose sum is -4 and product is -96, these are 8 and -12. Therefore, taking these two numbers back to our equation(x+8)(x-12)=0Therefore, x=-8 or 12Equally, using quadratic formula of [tex]x=\frac{-b±\sqrt{b^{2} -4ac}}{2a}[/tex] for equation [tex]ax^{2}+bx+c=0[/tex] and in our case a=1, b=-4 and c=-96[tex]x=\frac{4±\sqrt{(-4)^{2} -(4*1*-96)}}{2*1}[/tex]x=-8 or 12Since length can't be negative, hence x=12The length is 12 milesWidth=12-4=8 milesDimensions: 12 miles by 8 miles