# Give the correct answer and solve the given equation Let displaystyle{p}{left({x}right)}={2}+{x}{quadtext{and}quad}{q}{left({x}right)}={x}. Using the

Give the correct answer and solve the given equation
Let $p\left(x\right)=2+x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}q\left(x\right)=x$. Using the inner product find all polynomials $r\left(x\right)=a+bx\in P1\left(R\right)P$
(R) such that {p(x), q(x), r(x)} is an orthogonal set.
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mhalmantus
Notice that
$\left\{p,q\right\}={\int }_{-1}^{1}\left(2+x\right)xdx={\int }_{-1}^{1}2xdx+{\int }_{-1}^{1}{x}^{2}dx$
$={x}^{2}{|{}_{-{1}^{1}+}\frac{{x}^{3}}{3}|}_{-1}^{1}$
$=\left({1}^{2}-{\left(-1\right)}^{2}\right)+\left(\frac{{1}^{3}}{3}-\frac{{\left(-1\right)}^{3}}{3}\right)$
$=\left(1-1\right)+\left(\frac{1}{3}-\frac{-1}{3}\right)$
=2/3 Thus,$\left\{p,q\right\}\ne 0$,
so the set {p(x), q(x), r(x)} cannot be orthogonal
There exist none such r(x)