SOLUTION: You are making a frame for a painting. The painting is 17in by 11in. You want the total area of the painting and frame to be 315in^2. The frame will be the same width on all sides.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: You are making a frame for a painting. The painting is 17in by 11in. You want the total area of the painting and frame to be 315in^2. The frame will be the same width on all sides.      Log On


   

Question 872809: You are making a frame for a painting. The painting is 17in by 11in. You want the total area of the painting and frame to be 315in^2. The frame will be the same width on all sides. What should the outer dimensions of the frame be?
Found 2 solutions by josgarithmetic, mananth:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
The frame width is needed and should be found first. Let x be this width.

%2817%2B2x%29%2811%2B2x%29=Area inside the outer dimensions of the frame plus picture.
%282x%2B17%29%282x%2B11%29=315, total area of painting plus frame.
4x%5E2%2B34x%2B22x%2B17%2A11=315
4x%5E2%2B56x=128
highlight_green%28x%5E2%2B14x-32=0%29
%28x-2%29%28x%2B16%29=0
-
The realistic value for x is 2.
highlight%28x=2%29
-
Dimensions of the frame on the outside are 21 and 15 inches.

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Let the width be x
..
Length of picture 17 in
width of picture 11 in

Area = 187 In^2
Area of picture +frame 315 In^2
length of picture +frame 17 + 2 x
width of picture +frame 11 + 2 x

( 17 + 2 x ) ( 11 + 2 x ) = 315

187 + 34 x + 22 x + 4 X^2 = 315
4 X^2 + 56 x -128 = 0


4x^2+64x-8x-128=0
4x(x+16)-8(x+16)=0
(x+16)(4x-8)=0
x=-16 OR x=2
taking positive value
the outside dimensions are
19 by 13 in