numbers are x and y
x is 5 more than y
x=5+y
product is -3
xy=-3
solve
sub 5+y for x since x=5+y
(5+y)y=-3
5y+y^2=-3
add 3 to both sides
y^2+5y+3=0
use quadratic formula where
ay^2+by+c=0
y=[tex] \frac{-b+/- \sqrt{b^2-4ac} }{2a} [/tex]
a=1
b=5
c=3
y=[tex] \frac{-(5)+/- \sqrt{5^2-4(1)(3)} }{2(1)} [/tex]
y=[tex] \frac{-5+/- \sqrt{25-12} }{2} [/tex]
y=[tex] \frac{-5+/- \sqrt{13} }{2} [/tex]
y=[tex] \frac{-5+ \sqrt{13} }{2} [/tex] or [tex] \frac{-5- \sqrt{13} }{2} [/tex]
the second number is 5 more or 10/2 more
add 10/2 to each
x=[tex] \frac{5+ \sqrt{13} }{2} [/tex] or [tex] \frac{5- \sqrt{13} }{2} [/tex]
the numbers are
[tex] \frac{-5+ \sqrt{13} }{2} [/tex] and [tex] \frac{5+ \sqrt{13} }{2} [/tex] or
[tex] \frac{-5- \sqrt{13} }{2} [/tex] and [tex] \frac{5- \sqrt{13} }{2} [/tex]
aprox
the numbers are
-0.697224 and 5.967224 or
-4.302775 and 0.697224
