x+y=16
xy=12
x+y=16
subtract x from both side
y=16-x
subsitute for y
x(16-x)=12
16x-x^2=12
subtract add x^2 to both sides
16x=12+x^2
subtract 16x from both sides
x^2-16x+12=0
try and factor
can't
use quadratic equation which is
x=[tex] \frac{-b+/- \sqrt{b^{2}-4ac} }{2a} [/tex] when you have
ax^2+bx+c=0
x^2-16x+12=0
a=1
b=-16
c=12
subsitute
x=[tex] \frac{-(-16)+/- \sqrt{(-16)^{2}-4(1)(12)} }{2(1)} [/tex]
x=[tex] \frac{16+/- \sqrt{256-48} }{2} [/tex]
x= [tex] \frac{16+/- \sqrt{208} }{2} [/tex]
x=[tex] \frac{16+/- 4\sqrt{13} }{2} [/tex]
x=[tex] 8+/- 2\sqrt{13}[/tex]
x=[tex] 8- 2\sqrt{13}[/tex] or [tex] 8+ 2\sqrt{13}[/tex]
the answer is
the numbers are [tex] 8- 2\sqrt{13}[/tex] and [tex] 8+ 2\sqrt{13}[/tex] or
aprox 0.788897 and 15.2111
