Question 909293
A------------M-------------B

{{{M}}} is the midpoint => {{{AM=BM}}}

if {{{AM =x^2-6}}}
and {{{BM =x}}}

then {{{x^2-6=x}}} ...solve for {{{x}}}

{{{x^2-x-6=0}}} ...write {{{-x}}} as {{{2x-3x}}}

{{{ x^2+2x-3x-6 = 0 }}} ...group 

{{{ (x^2+2x)-(3x+6) = 0 }}}

{{{ x(x+2)-3(x+2) = 0 }}}

{{{ (x-3)(x+2) = 0 }}}

solutions

if {{{ x-3= 0 }}} => {{{x=3}}}

if {{{ x+2= 0 }}} => {{{x=-2}}} since {{{AB}}} is a line and we need the length, we will use only positive solution {{{x=3}}}

now find {{{AM =x^2-6}}} and {{{BM =x}}}

{{{BM =x}}}=> {{{BM =3}}}

{{{AM =3^2-6}}} => {{{AM =3}}} 


then {{{AB=AM+BM=3+3=6}}}