a^2 + b^2 = 20; ab + 6b = 32; what are all the pairs (a; b)

ab + 6b = 32    →  b(a + 6)  = 32  →   b = 32/(a + 6)   (2)

Substitue (2) into (1) and we  have

a^2 + [32/(a + 6)]^2  = 20      simplify

a^2(a + 6)^2 + 1024 = 20(a+6)^2

a^2(a^2 + 12a + 36) + 1024 = 20a^2 + 240a + 720

a^4 + 12a^3 + 36a^2 + 1024 = 20a^2 + 240a + 720

a^4 + 12a^3 + 16a^2 - 240a + 304  = 0

Using the on site solver we have one real soution for a

$${{\mathtt{a}}}^{{\mathtt{4}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}}{\mathtt{\,\times\,}}{{\mathtt{a}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{16}}{\mathtt{\,\times\,}}{{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{240}}{\mathtt{\,\times\,}}{\mathtt{a}}{\mathtt{\,\small\textbf+\,}}{\mathtt{304}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{8}}\\
{\mathtt{a}} = {\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{8}}\\
{\mathtt{a}} = {\mathtt{2}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{3.464\: \!101\: \!615\: \!137\: \!754\: \!6}}{i}\\
{\mathtt{a}} = {\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3.464\: \!101\: \!615\: \!137\: \!754\: \!6}}{i}\\
{\mathtt{a}} = {\mathtt{2}}\\
\end{array} \right\}$$

And b =

32 /(a + 6)   = 32 /(2 + 6)  =  32 /8   = 4

You can do this on the calculator here using the "Equation" mode:

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Thanks Alan and CPhill    

A very nice display from each of you.:)