Respuesta :
Answer:
12.808 meter is the length should Rickey use for the mural.
Step-by-step explanation:
Area of the mural = A = [tex]50 m^2[/tex]
Length of the mural = l
Width of the mural = w
l = 5 + 2w
Area of the rectangle = l × w
[tex]A=(5+2w)\times w[/tex]
[tex]50 m^2=5w+2w^2[/tex]
[tex]2w^2+5w-50=0[/tex]
Solving above equation with the help of Completing squares method:
[tex](\sqrt{2}w)^2+2\times \sqrt{2}w\times \frac{5}{2\sqrt{2}}+(\frac{5}{2\sqrt{2}})^2-50=(\frac{5}{2\sqrt{2}})^2[/tex]
[tex](a+b)^2=a^2+b^2+2ab[/tex]
[tex](\sqrt{2}w+\frac{5}{2\sqrt{2}})^2=(\frac{5}{2\sqrt{2}})^2+50[/tex]
[tex](\sqrt{2}w+\frac{5}{2\sqrt{2}})^2=\frac{25}{8}+50[/tex]
[tex](\sqrt{2}w+\frac{5}{2\sqrt{2}})^2=\frac{425}{8}[/tex]
[tex](\sqrt{2}w+\frac{5}{2\sqrt{2}})=\pm\sqrt{\frac{425}{8}}[/tex]
w = 3.904 m (accept)
w = -6.404 m (reject)
l = (5 + 2w)m = 5 + 2 × (3.904) m= 12.808 m
12.808 meter is the length should Rickey use for the mural.
