Question 551052
An obvious solution is {{{k=0}}}.
However, {{{k=24}}} is another solution, giving both equations,
{{{x^2-11x+24=0}}} and {{{x^2-14x+48=0}}}
the common root {{{x=8}}}.
{{{graph(300,300,-2,13,-8,32,x^2-11x+24,x^2-14x+48)}}}
A common root would be a solution of the non-linear system of non-linear equations:
{{{x^2-11x+k=0}}}
{{{x^2-14x+2k=0}}}
We can combine them to give
{{{x^2-11x+k=x^2-14x+2k}}} --> {{{3x=k}}} --> {{{x=k/3}}}
which we could use, along with the friendliest of the quadratic equations above to search for all possible solutions.
{{{x^2-14x+2k=0}}} has the solutions
{{{x=7+-sqrt(49-2k)}}}
Combining them, we get
{{{k/3=7+-sqrt(49-2k)}}} --> {{{k=21 +- 3sqrt(49-2k)}}} --> {{{(k-21)^2=3^2(49-2k)}}} --> {{{k^2-42k+441=441-18x}}} --> {{{k^2-24k=0}}} --> {{{k(k-24)=0}}}
So we get two answers {{{k=0}}} and {{{k=24}}}, and both verify.
For k=0, we get x=0/3=0.
For k=24 we get x=24/3=8.
The solutions are the (x,k) pairs (0,0) and (8,24).