Question 824722
1) If the equations {{{ax^2+bx+c=0}}} and {{{bx^2+ax+c=0}}} ,
where {{{a<>b}}} and {{{c<>0}}} , have a common root, that root will satisfy
{{{ax^2+bx+c=bx^2+ax+c}}} .
 
Solving:
{{{ax^2+bx+c=bx^2+ax+c}}}
{{{ax^2+bx+cross(c)=bx^2+ax+cross(c)}}}
{{{ax^2+bx=bx^2+ax}}}
{{{ax^2-bx^2+bx-ax=0}}}
{{{x(ax-bx+b-a)=0}}}
If {{{x=0}}} were a solution of {{{ax^2+bx+c=0}}} , {{{c}}} would be zero:
{{{a*0^2+b*0+c=0+0+c=c=0}}} .
So we know that,
{{{system(x<>0,x(ax-bx+b-a)=0)}}} --> {{{(ax-bx+b-a)=0}}}
{{{(ax-bx+b-a)=0}}}
{{{(a-b)x+b-a=0}}}
{{{(a-b)x=a-b}}}
We can divide both sides of the equal sign by {{{(a-b)<>0}}} , to get
{{{highlight(x=1)}}}.
(We know that {{{(a-b)<>0}}} because {{{a<>b}}} ).
 
Since we have concluded that only {{{x=1}}} could be a solution to both equations,
substituting {{{1}}} for {{{x}}} must make both equations true.
Substituting into {{{ax^2+bx+c=0}}} ,
we get
{{{0=ax^2+bx+c=a*1^2+b*1+c}}}
{{{0=ax^2+bx+c=a+b+c}}}
{{{highlight(a+b+c=0)}}}
We get the same using the other equation.
The two equations could be {{{x^2-3x+2=0}}} , with roots {{{x=1}}} and {{{x=2}}} ,
and {{{-3x^2+x+2=0}}} , with roots {{{x=1}}} and {{{x=-2/3}}}.
 
NOTE:
The only solution that we found that could be a common solution to both equations was {{{x=1}}} .
There must be two more, different solutions, one for each equation.
It cannot be {{{x=1}}} as a double solution to both equations,
because that would require {{{a=b}}} .
If {{{x=1}}} were the only solution to both equations, the equations would be
{{{0=ax^2+bx+c=a(x-1)^2=a(x^2-2x+1)=ax^2-2ax+highlight(a)}}} , and
{{{0=bx^2+ax+c=b(x-1)^2=b(x^2-2x+1)=bx^2-2bx+highlight(b)}}} .
The independent term in both should be
{{{c=highlight(a)}}} and {{{c=highlight(b)}}} .
That would require {{{a=b}}} .
 
2) {{{(x-k)^ 2 +2x-k=0}}}
{{{(x-k)^ 2 +2x-k-k=-k}}}
{{{(x-k)^ 2 +2x-2k=-k}}}
{{{(x-k)^ 2 +2(x-k)=-k}}}
{{{(x-k)^ 2 +2(x-k)+1=1-k}}}
{{{((x-k) +1)^ 2=1-k}}}
So {{{1-k>=0}}} --> {{{1>=k}}} <--> {{{k<=1}}}
 
Alternatively,
{{{(x-k)^ 2 +2x-k=0}}}
{{{x^2-2kx+k^2+2x-k=0}}}
{{{x^2+(2-2k)x+k^2-k=0}}}
For real solutions we need the discriminant to be non negative
{{{(2-2k)^2-4*1*(k^2-k)>=0}}}
{{{4-8k+4k^2-4k^2+4k>=0}}}
{{{4-8k+4k>=0}}}
{{{4>=4k}}}
{{{1>=k}}} <--> {{{k<=1}}}