<s>[INST] Solving an equation with three quadratic radicals in the set of real numbers How do we solve the following equation in the set of real numbers?
$$(26-x)\cdot\sqrt{5x-1} -(13x+14)\cdot\sqrt{5-2x} + 12\sqrt{(5x-1)\cdot(5-2x) }= 18x+32.$$
I tried putting $a=\sqrt{5x−1}$ and $b=\sqrt{5−2x}$ and then $2a^2+5b^2=23$.
 [/INST] This is another solution. 
Put $a=\sqrt{5x-1}$ and $b=\sqrt{5-2x}$. 
We have
\begin{equation*}
26-x = \dfrac{47}{23}(5x - 1 ) + \dfrac{129}{23}(5-2x) =  \dfrac{47}{23}a^2 + \dfrac{129}{23}b^2,
\end{equation*}
\begin{equation*}
13x+14= \dfrac{93}{23}(5x - 1 ) + \dfrac{83}{23}(5-2x) =  \dfrac{93}{23}a^2 + \dfrac{83}{23}b^2,
\end{equation*}
\begin{equation*}
18x+32= \dfrac{154}{23}(5x - 1 ) + \dfrac{178}{23}(5-2x) =  \dfrac{154}{23}a^2 + \dfrac{178}{23}b^2.
\end{equation*}
The given equation become 
\begin{equation*}
\begin{cases}
47a^3+129ab^2-93ba^2-83b^3-154a^2-178b^2+276ab = 0,\\
2a^2+5b^2=23.
\end{cases}
\end{equation*}
This system of equations is solved here
How do we solve the system of equations?
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