Question 697467
two integers differ by 8: x-y = 8


solve for y:  y = x-8


their product is -15:  xy = -15


plug in y = x-8:   x(x-8) = -15


Now solve for x


x(x-8) = -15


x^2 - 8x = -15


x^2 - 8x + 15 = 0


Use the quadratic formula to solve for x


{{{x = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{x = (-(-8)+-sqrt((-8)^2-4(1)(15)))/(2(1))}}} Plug in {{{a = 1}}}, {{{b = -8}}}, {{{c = 15}}}


{{{x = (8+-sqrt(64-(60)))/(2)}}}


{{{x = (8+-sqrt(4))/2}}}


{{{x = (8+sqrt(4))/2}}} or {{{x = (8-sqrt(4))/2}}}


{{{x = (8+2)/2}}} or {{{x = (8-2)/2}}}


{{{x = 10/2}}} or {{{x = 6/2}}}


{{{x = 5}}} or {{{x = 3}}}


Now that you know x = 5, you can use this to find that y = -3


Or if x = 3, then you can find out that y = -5. Both of these y values are found when you plug in the corresponding x value into {{{y = x - 8}}} and you solve for y.


So the two numbers are 5 and -3. Also, another pair of numbers is 3 and -5