Question 985353
Find two numbers whose sum is 19 and the product is 48
<pre>
{{{system(x+y=19,
xy=48)}}}

y=19-x

xy=48
x(19-x)=48
19x-x² = 48
-x²+19x-48 = 0

a=-1, b=19, c=-48

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

{{{x = (-(19) +- sqrt( (19)^2-4*(-1)*(-48) ))/(2*(-1)) }}}

{{{x = (-19 +- sqrt(361-192 ))/(-2) }}}

{{{x = (-19 +- sqrt(169 ))/(-2) }}}

{{{x = (-19 +- 13)/(-2) }}}

Using the +

{{{x = (-19 + 13)/(-2) }}}

{{{x = (-6)/(-2) }}}

{{{x = 3}}}

{{{x = 3}}}, 
{{{y=19-x}}}
{{{y=19-3}}}
{{{y=16}}}
      
So 3+16 = 19,  3*16 = 48

Using the -

{{{x = (-19 - 13)/(-2) }}}

{{{x = (-32)/(-2) }}}

{{{x = 16}}}

{{{x = 16}}}, 
{{{y=19-16}}}
{{{y=19-16}}}
{{{y=3}}}
      
So 16+3 = 19,  16*3 = 48

Not really a different answer, just swapping the letters.

Answer: 3 and 16

Edwin</pre>