Gaussian Quadrature - Construction Suppose you have $w(x)= 1/\sqrt{x}$ as your weight function, and the integration of the form $\int_0^1 f(x) w(x) dx$.  I am tasked with creating a quadrature of exactness 3. So I know I need a polynomial with degree $n=2$, and that I need to map the interval to the interval $[-1,1]$. 
So far, I think I need to use a Lagrange polynomial degree two, so I decided to use $p_2(x)=x^2 -1/3$. I can remap it as follows:
$$\int_0^1 f(x) \, dx =\int_{-1}^1 f\left( \frac{t + 1}{2} \right) \frac{1}{2} \, dt$$
The error term then should be 
$$\frac{f^{(4)}(\xi)}{4!} \int_0^1 p_2^2(x) w(x) \, dx$$
And this is the point where I get stuck. So I have the polynomial $p_2$, and I know how to remap it to the right interval, but how do I go on from here to construct the quadrature?
 A: The plan is to use Gram-Schmidt to create orthogonal polynomials, as is described in the answer here.  In that question, the weighting function was $w(x)=\frac{1}{1+x^2}$ but the mechanics are the same.
Starting with $p_0(x)=1$ we can calculate the next two iteratively
$$
p_1(x)=x-\frac{1}{3}\\
p_2(x)=x^2-\frac{6}{7}x +\frac{3}{35}.
$$
To achieve an exact quadrature for third order polynomials, this is all that is needed.  For $N-1$ Gaussian type quadrature, the abscissa are chosen to be the roots of the $N^\textrm{th}$ orthogonal polynomial.  In this case, $N=2$ and the abscissa are $x_i\in \left\{\frac{2}{35}\sqrt{30} \mp \frac{3}{7} \right\}$.
Armed with the abscissa and polynomials, the next step is to solve the tower equations to determine the weights
$$
\sum_{i=0}^1 p_n(x_i)w_i = 2\delta_{n,0}
$$
where the constant $2$ came from the normalization of $p_0(x)$, $\int_0^1\frac{p_0^2(x)}{\sqrt{x}}dx  = \int_0^1\frac{1}{\sqrt{x}}dx=2$
This leads to the matrix equation
$$
\left[\begin{array}{cc}
1 & 1 \\
\frac{2}{21}-\frac{2}{35}\sqrt{30} & \frac{2}{21}+\frac{2}{35}\sqrt{30}
\end{array}\right]
\left[\begin{array}{c}
w_0\\
w_1
\end{array}\right] =\left[\begin{array}{c}
2\\
0
\end{array}\right]
$$
which has solution $w_i\in \left[1\pm\frac{1}{18}\sqrt{30} ​\right]$.
This was verified by a small Sagemath (iPython) routine:
def gqsqrt(f):
    xi = vector([-2*sqrt(30)/35 + 3/7, 2*sqrt(30)/35 + 3/7])
    wi = vector([1+sqrt(30)/18,1-sqrt(30)/18])
    fncall = vector([f(x) for x in xi])
    return expand(wi.dot_product(fncall))

