Question 818245
The solutions to {{{ax^2+bx+c=0 }}} are {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} ,
which I will temporarily abbreviate as {{{x = (-b +- sqrt(DELTA))/(2*a) }}}
with {{{DELTA=b^2-4*a*c}}} .
 
If {{{(-b+sqrt(DELTA))/(2*a)= 2((-b-sqrt(DELTA))/(2*a))}}} ,
{{{-b+sqrt(DELTA)= 2(-b-sqrt(DELTA))}}}
{{{-b+sqrt(DELTA)= -2b-2sqrt(DELTA)}}}
{{{3sqrt(DELTA)= -b}}} which squares into {{{9DELTA=b^2}}} .
 
If {{{(-b-sqrt(DELTA))/(2*a)= 2((-b+sqrt(DELTA))/(2*a))}}} ,
{{{-b-sqrt(DELTA)= 2(-b+sqrt(DELTA))}}}
{{{-b-sqrt(DELTA)= -2b+2sqrt(DELTA)}}}
{{{-b=3sqrt(DELTA)}}} which squares into {{{b^2=9DELTA}}} .
 
Either way {{{b^2=9DELTA}}} which means {{{b^2=9(b^2-4*a*c)}}} .
So {{{b^2=9b^2-36*a*c)}}} , and adding {{{36*a*c-b^2}}} to both sides we get
{{{36*a*c=8b^2)}}} , which dividing both sides by 4 gives us
{{{9ac=2b^2)}}} or {{{highlight(2b^2=9ac)}}} .