Question 1049902

let numbers be {{{x}}} and {{{y}}}

if the product of two numbers is {{{96}}}, we have

{{{xy=96}}}........solve for {{{x}}}

{{{x=96/y}}}...............eq.1

if one number is {{{4}}} more than the other number, we have

{{{x=y+4}}}.........eq.2

from eq.1 and eq.2 we have

{{{96/y=y+4}}}........solve for {{{y}}}

{{{96=y(y+4)}}}

{{{96=y^2+4y}}}

{{{y^2+4y-96=0}}}.............use quadratic formula


{{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}...since {{{a=1}}},{{{b=4}}}, and {{{c=-96}}} we have


{{{y = (-4 +- sqrt( 4^2-4*1*(-96) ))/(2*1) }}}

{{{y = (-4 +- sqrt( 16+384 ))/2 }}}

{{{y = (-4 +- sqrt( 400 ))/2 }}}

{{{y = (-4 +- 20)/2 }}}

{{{y = (-cross(4)2 +- cross(20)10)/cross(2)1 }}}

{{{y = (-2 +- 10) }}}

solutions:

{{{y = -2 + 10 }}}->{{{y = 8 }}}
or
{{{y = -2- 10 }}}->{{{y = -12 }}}

now find corresponding {{{x}}} for each {{{y}}}

{{{x=y+4}}}.........eq.2...if {{{y = 8 }}}

{{{x=8+4}}}
{{{x=12}}}

{{{x=y+4}}}.........eq.2...if {{{y = -12 }}}

{{{x=-12+4}}}
{{{x=-8}}}


so, we have two solutions:

1.{{{x=12}}} and {{{y = 8 }}}
2.{{{x=-8}}} and  {{{y = -12 }}}