Given:
Width of the rectangle=
[tex]w[/tex]
Length of a rectangle=
[tex]l=w+4[/tex]
Area of the rectangle=
[tex]A=50[/tex]
Required:
Width of the rectangle.
Answer:
Area of the rectangle=
[tex]\begin{gathered} A=l\times w \\ 50=(w+4)\times w \\ 50=w^2+4w \\ w^2+4w-50=0 \\ w=\frac{-4\pm\sqrt{16-4(1)(-50)}}{2} \\ w=\frac{-4\pm\sqrt{16+120}}{2} \\ w=-2+3\sqrt{6,}w=-2-3\sqrt{6} \end{gathered}[/tex]
Solving this equation by quadratic equation method we get,
[tex]w=-2+3\sqrt{6,}w=-2-3\sqrt{6}[/tex]
Final Answer:
The width of the rectangle is,
[tex]w=-2+3\sqrt{6,}w=-2-3\sqrt{6}[/tex]
