Find $x_0$ when a 3x3 symmetric matrix has equal eigenvalues The question goes like this: There is a symmetric matrix:$$A=\begin{bmatrix}3 & 0 & 0\\ 0 & x & 2\\0 & 2 & x\end{bmatrix}$$
Find the value(s) of $x$ for which $A$ has at most two distinct eigenvalues. (Eigenvalues like $3,2,2$)
In my attempts to solve this problem, I got the characteristic equation as:
$$\lambda^3-(2x+3)\lambda^2+(x^2+3x-2)\lambda-3(x^2-4)=0$$
I am unable to proceed any further than this. Should I try to solve for $\lambda$ by putting appropriate values in the equation, then find $x$? 
Is there any property that I seem to be missing?
 A: you should not have solved all you parentheses
you  had $$(3-\lambda)*[(x-\lambda)^2-4)]$$
so $$(3-\lambda)=0$$ is one solution  and $$[(x-\lambda)^2-4)]=0$$ the next.
A: Observe that 
$$A\begin{bmatrix}1\\0\\0\end{bmatrix}=3\begin{bmatrix}1\\0\\0\end{bmatrix}.$$
Thus $\lambda=3$ is an eigenvalue of this matrix. 
Also observe
$$A\begin{bmatrix}0\\1\\1\end{bmatrix}=(x+2)\begin{bmatrix}0\\1\\1\end{bmatrix}.$$
Thus $\lambda=x+2$ is an eigenvalue of this matrix as well. 
Now the sum of the eigenvalues is the trace of the matrix. Let the other eigenvalue be $\lambda_3$, then
$$3+(x+2)+\lambda_3=2x+3 \implies \color{red}{\lambda_3=x-2}.$$
So the three eigenvalues are $\boxed{3,x+2}$ and $\boxed{x-2}$. We want at most two distinct eigenvalues. Observe that when $x=1,5$ then two of them are equal, hence only two distinct eigenvalues.
When $\color{red}{x=1}$, the eigenvalues are $\color{blue}{3,3,-1}$.
When $\color{red}{x=5}$, the eigenvalues are $\color{blue}{3,7,3}$.
When $\color{red}{x \neq 1,5}$, the eigenvalues are all $\color{blue}{\text{distinct}}$.
For no value of $x$ can all the eigenvalues be the same.

Further addition to the solution:
In case you are not aware of the trace result, you can still get the third eigenvalue by observing that
$$A\begin{bmatrix}0\\1\\-1\end{bmatrix}=(x-2)\begin{bmatrix}0\\1\\-1\end{bmatrix}.$$
Thus $\lambda=x-2$ is an eigenvalue of this matrix as well. 
