<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] I solved by another way. This is my solution. Please comment to me. Put $t=\sqrt{5x-1}-\sqrt{5-2x}.$ 
We have
$$t^2=3x+4-2\sqrt{(5x-1)( 5-2x)}$$ 
and 
$$t^3=(14-x)\sqrt{5x-1}-(13x+2)\sqrt{5-2x}.$$
And then, the given equation has the form
$(t - 2)^3 = 0$, that is mean $t = 2$. With $t = 2, $ we have
\begin{equation*}
\sqrt{5x-1}-\sqrt{5-2x} = 2.
\end{equation*}
This equation has a solution $x =2.$
Thus, the given equation has the only root $x = 2.$
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