 Solution

Write $p=a+bx$. orthogonality of the residual to $1,x$ givesThe first integral follows from $t=1-x^2$; for the second use $x=\sin \theta$, giving ${\int }_0^{\pi /2}{\sin }^2\theta {\cos }^2\theta d\theta =\pi /16$. The coefficient matrix has determinant $1/72$, so solving yieldsPart (a) proves this is the unique best weighted $L^2$ approximation. The weight $x$ is essential; using ordinary unweighted normal equations would give different coefficients.