Answer :
the length of a rectangle is 11 yd more than twice the width and the area of the rectangle is 63 yd squared. find the dimensions of the rectangle.
Let
L ------> the lenght
W ----> the width
we know that
the area of rectangle is
A=L*W
A=63 yd2
63=L*W -------> equation 1
and
L=2W+11 ------> equation 2
substitute equation 2 in equation 1
63=(2W+11)*w
2W^2+11w-63=0
solve the quadratic equation using the formula
a=2
b=11
c=-63
substitute
[tex]w=\frac{-11\pm\sqrt[]{11^2-4(2)(-63)}}{2(2)}[/tex][tex]\begin{gathered} w=\frac{-11\pm\sqrt[]{625}}{4} \\ \\ w=\frac{-11\pm25}{4} \\ \end{gathered}[/tex]the solutions for W are
w=3.5 and w=-9 (is not a solution, because is negative)
so
Find the value of L
L=2W+11 -------> L=2(3.5)+11
L=18
therefore
the dimensions are