Question 1109206: a number is one less than twice another number,and the difference between their squares is 85. find the numbers Found 2 solutions by ankor@dixie-net.com, greenestamps:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! a number is one less than twice another number,
a = 2b - 1
and the difference between their squares is 85.
a^2 - b^2 = 85
replace a with (2b-1)
(2b-1)^2 - b^2 = 85
FOIL (2b-1)(2b-1)
4b^2 - 2b - 2b + 1 - b^2 = 85
Combine to form a quadratic equation on the left
3b^2 - 4b + 1 - 85 = 0
3b^2 - 4b - 84 = 0
you can use the quadratic formula; a=3; b=-4; c=-84, but this will factor
(3b+14)(b-6) = 0
The positive integer solution
b = 6
Find a
a = 2(6) - 1
a = 11
:
find the numbers 11, 6
:
:
Check:
11^2 - 6^2 =
121 - 36 - 85
The statement of the problem implies that the numbers are whole numbers. So solving the problem requires less work if you look for numbers for which the difference of the squares is 85 and find such a pair with one of the numbers being 1 less than twice the other.
There are two possibilities in whole numbers: ; --> (a,b) = (43,42) no.... ; --> (a,b) = (11,6) YES!!!
