Given on the problem are the following: A (Area), L( Length) and W (Width)
[tex]A = 160[/tex]
[tex]L = 2W + 3[/tex]
- The formula in finding the Area is :
[tex]A = L x W[/tex]
- Substitute L:
[tex]A = (2W + 3) W[/tex]
[tex]160 = 2W^{2} + 3W[/tex]
[tex] 2W^{2} +3W -160 = 0[/tex]
-Solve using quadratic equation:
[tex] ax^{2}+ bx + c = 0 [/tex]
- To solve for x ( which is the W)
[tex]x = \frac{-b \sqrt{ b^{2}-4ac } }{2a }
\\ \\
x = \frac{-3 \sqrt{ (3)^{2}-4(2)(-160) } }{2(2) }
\\ \\
x = \frac{-3 \sqrt{ 9+1280 } }{4 }
\\ \\
x = \frac{-3 \sqrt{ 1289 } }{4 }
\\ \\
x = \frac{-3 + 35.90 }{4 } or \frac{-3 - 35.90 }{4 }
\\ \\
x = 8.225 or-9.725
[/tex]
- The Width is the positive value of x so:
[tex]W = 8.225 m [/tex]
- The Length is:
[tex]L = 2(8.225)+3 [/tex]
[tex]L = 19.45 m[/tex]
