SOLUTION: A SQUARE POSTER HAS SIDES MEASURING 2 FEET LESS THAN THE SIDES OF A SQUARE SIGN. IF THE DIFFERENCE BETWEEN THEIR AREAS IS 20 SQUARE FEET, FIND LENGTHS OF THE SIDES OF THE POSTER AN
Algebra ->
Problems-with-consecutive-odd-even-integers
-> SOLUTION: A SQUARE POSTER HAS SIDES MEASURING 2 FEET LESS THAN THE SIDES OF A SQUARE SIGN. IF THE DIFFERENCE BETWEEN THEIR AREAS IS 20 SQUARE FEET, FIND LENGTHS OF THE SIDES OF THE POSTER AN
Log On
Question 372148: A SQUARE POSTER HAS SIDES MEASURING 2 FEET LESS THAN THE SIDES OF A SQUARE SIGN. IF THE DIFFERENCE BETWEEN THEIR AREAS IS 20 SQUARE FEET, FIND LENGTHS OF THE SIDES OF THE POSTER AND THE SIGN. Found 2 solutions by Greenfinch, solver91311:Answer by Greenfinch(383) (Show Source):
You can put this solution on YOUR website! side of poster is x
side of sign is x+2
(x + 2)^2 - x^2 = 20
x^2 + 4x +4 - x^2 = 20
4x + 4 = 20
x = 4
Check
poster is 16 sq ft
sign is 36 sq ft
Typing in all caps is the electronic communications equivalent of shouting. It is both annoying and rude.
Let represent the measure of the side of the larger square. Then is the measure of the side of the smaller square, is the area of the smaller square, and is the area of the larger square.
Solve for for the sides of the sign, subtract 2 for the sides of the poster.
John
My calculator said it, I believe it, that settles it
