Area = 300 square feetLength must be 12 feet longer than widthFind the dimension of the pen

Answer:[tex]Length=24.32[/tex] feet[tex]width=12.32[/tex] feetStep-by-step explanation:Let x be the width of a rectangleLet l be the length of a rectangleLet A be the area of rectangleGiven.Length must be 12 feet longer than widththerefore[tex]Length=12+w[/tex][tex]l=12+w[/tex]Area of rectangle [tex]A=300[/tex]We know that area of rectangle is [tex]A=l\times w[/tex]---------------------(1)put all known values in equation 1[tex]300=(12+x)\times x[/tex][tex]x^{2} +12x=300\\x^{2} +12x-300=0[/tex]Find roots by this formula[tex]=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex][tex]=\frac{-12\pm\sqrt{12^{2}-4(1)(-300)}}{2(1)}[/tex][tex]=\frac{-12\pm\sqrt{1344}}{2}[/tex][tex]x=-6+4\sqrt{21} \\x=-6-4\sqrt{21}[/tex][tex]x=12.32\\x=-24.32[/tex]check both value in equation 1for [tex]x=12.32[/tex][tex]A=(12+12.32)\times 12.32[/tex][tex]A=299.62[/tex][tex]x=12.32[/tex] is satisfy the equationSo The length of rectangle [tex]l=12+12.32[/tex]=24.32 feetAnd width is 12.32 feet