Given the quadrature rule $Q[f]=f(x_1)+f(x_2) \approx \int_{-1}^{1}f(x)dx = I[f]$ , find the nodes $x_1, x_2$ Given is the following quadrature rule :
$$Q[f]=f(x_1)+f(x_2) \approx \int_{-1}^{1}f(x)dx = I[f]$$ and the degree of accuracy $q \geq2$. We're asked to find the nodes $x_1, x_2$. 
May I ask you to give me feedback on my attempt ?

We evaluate $I[x^0]=2, I[x^1]=0, I[x^2]= \frac{2}{3}$. 
Thus: $$x^0: f(x_1)+f(x_2)= 1 + 1 \overset{!}{=} 2$$
$$x^1: f(x_1)+f(x_2)= x_1 + x_2 \overset{!}{=} 0$$
$$x^2: f(x_1)+f(x_2)= x_1^2 + x_2^2 \overset{!}{=} \frac{2}{3}$$
Thus: $x_2 = - x_1 \rightarrow x_1 = \pm \frac{1}{\sqrt{3}} \rightarrow x_2 = \pm \frac{1}{\sqrt{3}}$

Is what I did correct ? Did I forget something ? Thanks for your feedback !
 A: What you have done is fine, but for higher orders you might want to systematize the process via Newton's identities. In that case you have the sums of powers
$$p_k=\sum_{i=1}^nx_i^k$$
And the elementary symmetric polynomials $e_k$ which are the sums of all the distinct products of the $n$ variables taken $k$ at a time. Thus
$$\begin{align}e_0&=1\\
e_1&=\sum_{i=1}^nx_i\\
e_2&=\sum_{i=1}^{n-1}\sum_{j=i+1}^nx_ix_j\\
e_3&=\sum_{i=1}^{n-2}\sum_{j=i+1}^{n-1}\sum_{k=j+1}^nx_ix_jx_k\end{align}$$
And so on. Newton's identities read, for $n\ge k\ge1$,
$$ke_k=\sum_{i=1}^k(-1)^{i-1}e_{k-i}p_i$$
Thus in your case they read $p_0=2$, $p_1=0$, $p_2=\frac23$, $e_0=1$, $1e_1=e_1=e_0p_1=(1)(0)=0$, and $2e_2=e_1p_1-e_0p_2=(0)(0)-(1)\left(\frac23\right)=-\frac23$. Then the equation that $x_1$ and $x_2$ must satisfy is
$$\sum_{i=0}^n(-1)^ie_ix^{n-i}=x^2-(0)x+\left(-\frac13\right)=x^2-\frac13=0$$
So could you apply this procedure to find the corresponding Chebyshev quadrature rule with equal weights and $3$ nodes?
