Is there a basis for which the quadratic forms are both diagonalized? The question asks if there is a basis for which both of the following quadratic forms are in a diagonalized form: 
$$
q_1(x_1,x_2) = x_1^2 + x_1x_2 - x_2^2
$$
$$
q_2(x_1,x_2) = x_1^2 - 2x_1x_2
$$
What i tried:
I tried to see if one of the quadratic forms is positive definite, but I got that both are not. 
Yet, it's not enough to say that none are positive definite in order to prove that there isn't a basis for which both are diagonalized, or at least I am not familiar with such a sentence. 
So how can I determine this ? 
Thank you. 
 A: Let's translate the problem into the language of matrices. Write $q_1(x) = x^T A x$ and $q_2(x) = x^T B x$ with $A,B$ symmetric and assume that $q_1,q_2$ are simultaneously diagonalizable. Then we can find an invertible $P$ with
$$ P^TAP = D_1, P^TBP = D_2 $$
where $D_1,D_2$ are diagonal. Let's assume that $A,B$ (and so $D_1,D_2$) are invertible. Then
$$ A = \left( P^T \right)^{-1} D_1 P^{-1}, B = \left( P^T \right)^{-1} D_2 P^{-1} $$
so
$$ A^{-1} B = P D_1^{-1} P^T \left( P^T \right)^{-1} D_2 P^{-1} = P \left( D_1^{-1} D_2 \right) P^{-1} $$
which implies that $A^{-1}B$ is similar to a diagonal matrix $D_1^{-1} D_2$. So a necessary condition for $q_1,q_2$ to be simultaneously diagonalizable is for $A^{-1}B$ to be a diagonalizable matrix.

In your case,
$$ A = \begin{pmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & -1 \end{pmatrix}, B = \begin{pmatrix} 1 & -1 \\ -1 & 0 \end{pmatrix} $$
so
$$ A^{-1} B = \frac{1}{5} \begin{pmatrix} 2 & -4 \\ 6 & -2 \end{pmatrix}. $$
This matrix $5 A^{-1} B$ has characteristic polynomial $X^2 + 20$ which doesn't have real roots so it is not diagonalizable and hence your quadratic forms can't be simultaneously diagonalized over $\mathbb{R}$. 
A: The answer is no. 
In fact if there exists a base of $\mathbb{R}^2$ such that the symmetric matrixes $A,B$ associated to the two quadratic forms are diagonalizable in that base, then there exists an orthogonal change base matrix $P$ such that 
$P^TAP=I_1$
and 
$P^TBP=I_2$
where $I_1, I_2$ are diagonal matrixes. Then we get 
$AB=P^TI_1I_2 P=P^TI_2 I_1 P=BA$ 
because $P$ is orthogonal and so $P^{T}=P^{-1}$.
This means that you would have $AB=BA$. 
In your case 
$A=[[1, \frac{1}{2}], [\frac{1}{2}, -1]]$
$B=[[1, -1],[-1, 0]]$ 
and you can observe they not commute. 
A necessary condition to get two matrixes are simultaneously diagonalizable is that they commute. 
When the field is algebraically closed, then this condition is also sufficient. 
