SOLUTION: An aluminum brace used to support a wooden shelf has a length that is 1 inches more than twice the width of the shelf. The brace is anchored to the wall h inches below the shelf. F

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: An aluminum brace used to support a wooden shelf has a length that is 1 inches more than twice the width of the shelf. The brace is anchored to the wall h inches below the shelf. F      Log On


   

Question 1122945: An aluminum brace used to support a wooden shelf has a length that is 1 inches more than twice the width of the shelf. The brace is anchored to the wall h inches below the shelf. Find the width of the shelf and the length of the brace. (Assume h = 15.)
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
An aluminum brace used to support a wooden shelf has a length that is 1 inches more than twice the width of the shelf.
The brace is anchored to the wall h inches below the shelf.
Find the width of the shelf and the length of the brace. (Assume h = 15.)
:
draw this out, a right triangle
let w = the width of the shelf
then (2w+1) = the length of the brace (the hypotenuse)
w%5E2+%2B+15%5E2+=+%282w%2B1%29%5E2
w^2 + 225 = 4w^2 + 4w + 1
combine like terms to form a quadratic equation on the right
0 = 4w^2 - w^2 + 4w + 1 - 225
3w^2 + 4w -224 = 0
use the quadratic equation a=3; b=4; c=-224, but this will factor to
(3w+28)(w-8) = 0
the positive solution is all we want here
w = 8 inches is the width of the shelf
then
2(8) + 1 = 17 inches is the length of the brace
:
:
Check on your calc: 8^2 + 15^2 = 17^2