You can put this solution on YOUR website! One positive number exceeds another positive number by 5 .
The sum of their squares is 193.
find the numbers
:
x^2 + (x+5)^2 = 193
FOIL (x+5)(x+5)
x^2 + x^2 + 10x + 25 = 193
2x^2 + 10x + 25 - 193 = 0
2x^2 + 10x - 168 = 0
simplify, divide by 2
x^2 + 5x - 84 = 0
easily factors to
(x+12)(x-7) = 0
the positive solution
x = 7
:
:
Check: 7^2 + 12^2 =
49 + 144 = 193
x = y + 5 is one of your equations.
x^2 + y^2 = 193 is the other of your equations.
in the second equation, replace x with y + 5 to get:
(y + 5)^2 + y^2 = 193
simplify to get:
y^2 + 10y + 25 + y^2 = 193
combine like terms to get:
2y^2 + 10y + 25 = 193
subtract 193 from both sides of the equation to get:
2y^2 + 10y - 168 = 0
divide both sides of this equation by 2 to get:
y^2 + 5y - 84 = 0
factor to get:
(y + 12) * (y - 7) = 0
solve for y to get:
y = -12 or y = 7
value of y has to be positive, so y = 7 is a possible solution.
x = y + 5, therefore x = 12
your number appear to be 12 and 7.
12 - 7 = 5, which satisfies one of the requirements.
12^2 + 7^2 = 144 + 49 = 193, which satisfies the other requirement.
