I am trying to solve an equations such as the roots of $$k*x(11*x + 1) + d*x(11x + 1)$$ has to match the roots of this function $$x^2 + 0.1x + 6 + k*x(11*x + 1) + d*x(11x + 1)$$, where I have to solve values for k and d which match the criteria.
Is it even possible, I cannot seem to make mathmatica do it which make me question my method..
and What about for these for these
$$p*x == x^2 + 0.1x+(6+p)$$
or
$$kp*x (11 x^2 + 1 x) + ki (11 x + 1x) + kd*x^2 (11x + 1 x) == x (x^2 + 0.1 x + 6) + kp*x (11 x^2 + 1 x) + ki (11 x + 1 x) + kd*x^2 (11 x^2 + 1 x)$$
or
$$kp (11 x^2 + 1 x) + kd*x (11 x + 1 x) == x (x^2 + 0.1 x + 6) + kd*x(11 x^2 + 1 x)$$
one of them must be able to do so.
By factoring out $11x+1$ and $x$ the first function we find that it is equal to
$$ x (k+d)(11x+1), $$
so its roots are $x=0$ and $x = -1/11$.
By expanding the parentheses then collecting like powers of $x$ we find that the second function is equal to
$$ (1+11k+11d)x^2 + (k + d + 0.1) x + 6. $$
In order for this to have a root at $x = 0$ the constant term must be zero. Since it isn't, and since the constant term doesn't depend on $k$ or $d$ (it's just $6$), it doesn't have a root at $x = 0$ no matter what $k$ and $d$ are.