Question 197327
Find two real numbers:  x and y:
:
 that have a sum of 8
x + y = 8
y = (8-x); use for substitution
:
 and a product of 2.
x*y = 2
:
Replace y with (8-x) in the above equation:
x(8-x) = 2
:
8x - x^2 = 2
:
-x^2 + 8s - 2 = 0
:
Use the quadratic formula to find x
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this equation; a=-1; b=8; c=-2
{{{x = (-8 +- sqrt(8^2 - 4 * -1 * -2 ))/(2*-1) }}}
:
{{{x = (-8 +- sqrt(64 - 8 ))/(-2) }}}
:
{{{x = (-8 +- sqrt(56 ))/(-2) }}}
Two solutions:
{{{x = (-8 + 7.4833)/(-2) }}}
:
{{{x = (-.5167)/(-2) }}}
x = +.258; then y =7.742
and
{{{x = (-8 - 7.4833)/(-2) }}}
:
{{{x = (-15.4833)/(-2) }}}
x = +7.742; then y = .258
:
You can check these solutions in the original problem