Matrix eigenvalue and eigenvector question. 
Let $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \in K^{2 \times 2}$. If $b \neq 0$ and $\mu \in K$ is a root of the polynomial
  $$bx^2+(a-d)x+c=0 \, ,$$ then $a+b\mu$ is an eigenvalue of $A$ with
  eigenvector $\begin{bmatrix} 1 \\ \mu \end{bmatrix}$. If $b=0$, then
  $d$ is an eigenvalue of $A$ with eigenvector $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$.

After finding characteristic polynomial $p(x)=x^2-(a+d)x+(ad-bc)$,  I tried to find its roots but the steps got tedious. Where to use $\mu$ as a root of the given equation?  could you please give me some hints. Thank you.
 A: Case: $b=0$. Then the matrix is $A$ is triangular, so that directly one obtains $\text{char}_A(x)=(x-a)(x-d)$ whose roots are eigen values of $A$. Thus $d$ is an eigen value of $A$. Verify the eigen vector part.
Case: $b\ne 0$. I strongly feel that the polynomial should be $bx^2+(a-d)x-c=0$ instead of $+c$ OR $-bx^2+(a-d)x+c=0$ instead of $+b$. Please check this. Under this assumption let us call $\mu$ to be the positive root of the given polynomial. This means $$\mu=\dfrac{d-a+\sqrt{(a-d)^2+4bc}}{2b}$$
So $$a+b\mu=\dfrac{d+a+\sqrt{(a-d)^2+4bc}}{2}$$
which is precisely the positive root of the characteristic polynomial of $A$. Verify the eigen vectors.
Note: By positive root, I mean the root $\frac{-b+\sqrt{b^2-4ac}}{2a}$ which is just a terminology.
A: The simplest way to answer your question without computing the roots of the polynomial is as follows.
The solution of the characteristic polynomial are the eigenvalues. So if $(a+ \mu b)$ is an eigenvalue $p(a+ \mu b)=0$. Now $p(a+ \mu b)=b[b \mu²+\mu(a-d)-bc]$. However you wrote that $\mu$ is the root of the  polynomial $b \mu²+\mu(a-d)-bc$, then $b \mu²+\mu(a-d)-bc =0$ and $p(a+ \mu b)=0$.
