Answer:
The length of the rectangle is [tex]9\ cm[/tex]
Step-by-step explanation:
we know that
The area of rectangle is equal to
[tex]A=LW[/tex]
In this problem we have
[tex]A=27\ cm^{2}[/tex]
so
[tex]27=LW[/tex] ------> equation A
and
[tex]A=2w^{2}+3w[/tex]
so
[tex]27=2w^{2}+3w[/tex]
[tex]2w^{2}+3w-27=0[/tex]
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]2w^{2}+3w-27=0[/tex]
so
[tex]a=2\\b=3\\c=-27[/tex]
substitute in the formula
[tex]w=\frac{-3(+/-)\sqrt{3^{2}-4(2)(-27)}} {2(2)}[/tex]
[tex]w=\frac{-3(+/-)\sqrt{225}} {4}[/tex]
[tex]w=\frac{-3(+/-)15} {4}[/tex]
[tex]w=\frac{-3(+)15} {4}=3[/tex]
[tex]w=\frac{-3(-)15} {4}=-4.5[/tex]
The solution of the quadratic equation is
[tex]w= 3\ cm[/tex]
Find the value of L
[tex]27=L(3)[/tex]
[tex]L=27/3=9\ cm[/tex]
