SOLUTION: An Art dealer advises that for the best display of a painting, the area of the frame should be (1/5) the area of the painting. Determine the best width of the frame for a painting

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: An Art dealer advises that for the best display of a painting, the area of the frame should be (1/5) the area of the painting. Determine the best width of the frame for a painting       Log On


   

Question 130905: An Art dealer advises that for the best display of a painting, the area of the frame should be (1/5) the area of the painting. Determine the best width of the frame for a painting 20-cm by 2o-cm,if the width is to be the same on all sides. Round answer to nearest tenth.
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
20*20=400 IS THE AREA OF THE PICTURE
1:5=X:400
X=80 FOR THE AREA OF THE FRAME.
(20+2X)^2-400=80
400+80X+4X^2-400-80=0
4X^2+80X-80=0
4(X^2+40X-40)=0
USING THE QUADRATIC EQUATION WE GET:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
X=(-40+-SQRT[40^2-4*1*-40])/2*1
X=(-40+-SQRT[1,600+160])/2
X=(-40+-SQRT1,760)/2
X=(-40+-41.95)/2
X=(-40+41.95)/2
X=1.95/2
X=.975 ANSWER.
X=(-40-41.95)/2
X=-81.95/2
X=-40.975 ANSWER.