Question 957444


{{{ 3x^2+mx-12=0}}}.......use quadratic formula to find roots

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

{{{x = (-m+- sqrt( m^2-4*3*(-12) ))/(2*3) }}}

{{{x = (-m+- sqrt( m^2+144 ))/6 }}}  

{{{x = (1/6)(-m+- sqrt( m^2+144 )) }}} 

roots:


{{{x = (1/6)(-m+ sqrt( m^2+144 )) }}} 

or


{{{x = (1/6)(-m - sqrt( m^2+144 )) }}}


since given that their difference must be {{{5}}}, we can use it to solve for {{{m}}} 

{{{(1/6)(-m+ sqrt( m^2+144 ))-((1/6)(-m - sqrt( m^2+144 )))=5 }}}...both sides multiply by {{{6}}}

{{{6(1/6)(-m+ sqrt( m^2+144 ))-(6(1/6)(-m - sqrt( m^2+144 )))=5*6 }}}

{{{(-m+ sqrt( m^2+144 ))-(-m - sqrt( m^2+144 ))=30 }}}

{{{-cross(m)+ sqrt( m^2+144 )+cross(m) + sqrt( m^2+144 )=30 }}}

{{{ 2sqrt( m^2+144 )=30 }}}

{{{ sqrt( m^2+144 )=30/2 }}}

{{{ sqrt( m^2+144 )=15 }}}...............square both sides

{{{ (sqrt( m^2+144 ))^2=15^2 }}}

{{{  m^2+144 =225 }}}

{{{  m^2 =225-144 }}}

{{{  m^2 =81 }}}

{{{  m =sqrt(81) }}}

{{{  m =9 }}}

solutions: {{{  m =9 }}} or {{{  m =-9 }}}