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put this solution on YOUR website!A photo is 3 inches longer than it is wide. a 2-inch border is placed around the photo making the total area of the photo and border 108in^2.
What are the dimensions of the photo
:
Let x = the width of the photo
It says it is 3" longer than it is wide, therefore
(x+3) = the length of the photo
:
a two inch border will add 4 inches to the photo dimensions, therefore
(x+4) = width including the border
and
(x+3) + 4
which is
(x+7) = length including the border
:
Given that the area of the whole thing as 108 sq/in, therefore
(x+4) * (x+7) = 192
FOIL
x^2 + 7x + 4x + 28 = 192
:
x^2 + 11x + 28 - 192 = 0
:
x^2 + 11x - 164 = 0; our old friend, the quadratic equation!
:
Use the quadratic formula:
in this problem: a=1; b=11; c=-164
:
:
:
Positive solution
x =
x = 8.437" is the width of the photo
then
8.437 + 3 = 11.437" is the length
;
:
Check solution by finding the overall area
(8.437+4) * (11.437+4)) = 191.99 ~ 192
