Question 770348
What adds to make 5 and multiplies to get -3
<pre>
Let x and y be the answers:

x + y = 5
   xy = -3

Solve the first for y

    y = 5 - x

Substitute it in the second

   xy = -3

x(5 - x) = -3

5x - x² = -3

0 = x² - 5x - 3

Use the quadratic formula since that doesn't factor:

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

x = {{{(-(-5) +- sqrt( (-5)^2-4*(1)*(-3) ))/(2*(1)) }}}

x = {{{(5 +- sqrt(25+12 ))/2 }}}

x = {{{(5 +- sqrt(37))/2 }}}

That's two answers:

 x = {{{(5 + sqrt(37))/2 }}} and x = {{{(5 - sqrt(37))/2 }}}

Substituting in y = 5 - x

 y = 5 - {{{(5 + sqrt(37))/2 }}}  and y = 5 - {{{(5 - sqrt(37))/2 }}} 

y = {{{10/2}}} - {{{(5 + sqrt(37))/2 }}}  and y = {{{10/2}}} - {{{(5 - sqrt(37))/2 }}} 

     y = {{{(5 - sqrt(37))/2 }}}  and y = {{{(5 + sqrt(37))/2 }}}

So there is really only one solution since 
those just reverse the roles of x and y.

Answer: {{{(5 + sqrt(37))/2 }}}  and  {{{(5 - sqrt(37))/2 }}}

Checking:

Add them:

{{{(5 + sqrt(37))/2 }}}{{{""+""}}}{{{(5 - sqrt(37))/2 }}} {{{""=""}}}{{{(10 + sqrt(37)-sqrt(37))/2 }}} {{{""=""}}}  {{{10/2}}} {{{""=""}}} {{{5}}}.

Multiply them:

{{{(5 - sqrt(37))/2 }}}{{{""*""}}}{{{(5 + sqrt(37))/2 }}} {{{""=""}}} {{{(25 - 37)/4}}} {{{""=""}}} {{{(-12)/4}}} {{{""=""}}} {{{-3}}}.

Edwin</pre>