Question 739932


1 number {{{x}}} is equal to the square of another {{{y}}}. 

 {{{x=y^2}}}......eq.1


Find the numbers if both are positive and their sum is {{{x+y=1980}}}......eq.2


 {{{x=y^2}}}......eq.1
{{{x+y=1980}}}......eq.2
__________________________ solve the system

{{{x+y=1980}}}......eq.2...substitute {{{x}}} from eq.1

{{{y^2+y=1980}}}......eq.2

{{{y^2+y-1980=0}}}.....use quadratic formula


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


{{{y = (-1 +- sqrt( 1^2-4*1*(-1980) ))/(2*1) }}}


{{{y = (-1 +- sqrt( 1+7920 ))/2 }}}


{{{y = (-1 +- sqrt( 7921 ))/2 }}}


{{{y = (-1 +- 89)/2 }}}

we need only positive solution because given the numbers are both positive


{{{y = (-1 + 89)/2 }}}

{{{y =  88/2 }}}

{{{highlight(y =  44) }}}


find {{{x}}}


{{{x=y^2}}}......eq.1

{{{x=44^2}}}

{{{highlight(x=1936)}}}


check their sum:


{{{x+y=1980}}}......eq.2

{{{1936+44=1980}}}

{{{1980=1980}}}