Answer:
[tex]Length=24.32[/tex] feet
[tex]width=12.32[/tex] feet
Step-by-step explanation:
Let x be the width of a rectangle
Let l be the length of a rectangle
Let A be the area of rectangle
Given.
Length must be 12 feet longer than width
therefore
[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 1
for [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 equation
So The length of rectangle [tex]l=12+12.32[/tex]=24.32 feet
And width is 12.32 feet
