Answer:
[tex]w=-1+\sqrt{7}\ ft[/tex]
[tex]L=1+\sqrt{7}\ ft[/tex]
Step-by-step explanation:
we know that
The area of a rectangle is equal to
[tex]A=LW[/tex]
In this problem we have
[tex]A=6\ ft^{2}[/tex]
so
[tex]6=LW[/tex] ------> equation A
[tex]L=W+2[/tex] -----> equation B
substitute equation B in equation A
[tex]6=(W+2)W[/tex]
[tex]6=W^{2} +2W\\ \\W^{2} +2W-6=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]W^{2} +2W-6=0[/tex]
so
[tex]a=1\\b=2\\c=-6[/tex]
substitute in the formula
[tex]w=\frac{-2(+/-)\sqrt{2^{2}-4(1)(-6)}} {2(1)}[/tex]
[tex]w=\frac{-2(+/-)\sqrt{28}}{2}[/tex]
[tex]w=\frac{-2(+/-)2\sqrt{7}}{2}[/tex]
[tex]w=\frac{-2(+)2\sqrt{7}}{2}=-1+\sqrt{7}[/tex] ------> the value positive is the solution
[tex]w=\frac{-2(-)2\sqrt{7}}{2}=-1-\sqrt{7}[/tex]
therefore
[tex]w=-1+\sqrt{7}\ ft[/tex]
Find the value of L
[tex]L=W+2[/tex] ------> [tex]L=-1+\sqrt{7}+2=1+\sqrt{7}\ ft[/tex]
