Question 1132708
If the roots are ax^2+bx+c=0, differ by 1,
<pre>
Let the roots be r and r+1

{{{ar^2+br+c=0}}} and {{{a(r+1)^2+b(r+1)+c=0}}}

So {{{ar^2+br+c=a(r+1)^2+b(r+1)+c}}}

{{{ar^2+br+c=a(r^2+2r+1)^2+br+b+c}}}
 
{{{ar^2+br+c=ar^2+2ar+a+br+b+c}}}

{{{0=2ar+a+b}}}

{{{2ar+a+b=0}}}

{{{2ar= -a-b}}}

{{{r = (-a-b)/(2a)}}}   <-- that's one root

The other root is r+1, so we add 1 to both sides:

{{{r+1 = (-a-b)/(2a)+1}}}

{{{r+1 = (-a-b)/(2a)+(2a)/(2a)}}}

{{{r+1 = (-a-b+2a)/(2a)}}}

{{{r+1 = (a-b)/(2a)}}}  <-- that's the other root
</pre>
and prove that b^2=a(a+4c)
<pre>
Substitute either root in the original.  I'll use

{{{(a-b)/(2a)}}}, since it has only one - sign. 

{{{a((a-b)/(2a))^2+b((a-b)/(2a))+c=0}}}

{{{a((a^2-2ab+b^2)/(4a^2))+((ab-b^2)/(2a))+c=0}}} 

{{{cross(a)((a^2-2ab+b^2)/(4a^cross(2)))+((ab-b^2)/(2a))+c=0}}}

{{{((a^2-2ab+b^2)/(4a))+((ab-b^2)/(2a))+c=0}}}

Multiply through by 4a

{{{(a^2-2ab+b^2)+(2ab-2b^2)+4ac=0}}}

{{{a^2-2ab+b^2+2ab-2b^2+4ac=0}}}

{{{a^2-b^2+4ac=0}}}

{{{a^2+4ac=b^2}}}

{{{a(a+4c)=b^2}}}

{{{b^2=a(a+4c)}}}

Edwin</pre>