<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] If you put $a=\sqrt{5x-1}$ and $b = \sqrt{5-2x}$, the equation says
$$ (26-x) a - (13x+14) b + 12 a b = 18 x + 32$$
The resultant of $(26-x) a - (13x+14) b + 12 a b - 18 x - 32$ and $5x-1-a^2$ with respect to $a$ is
$$ -169\,{x}^{2}{b}^{2}+5\,{x}^{3}-588\,{x}^{2}b+356\,x{b}^{2}-585\,{x}^{
2}+1808\,xb-340\,{b}^{2}+2280\,x-1520\,b-1700
$$
The resultant of this and $5 - 2 x - b^2$ with respect to $b$ is $(49 x - 10)^3 (x - 2)^3$.
So $x = 10/49$ or $2$.  But now we have to check those by plugging in to the original equation.  $x=2$ does work but $10/49$ doesn't work: in fact it would give you
$$ - \left( 26-x \right) \sqrt {5\,x-1}+ \left( 13\,x+14 \right) \sqrt {5
-2\,x}+12\,\sqrt { \left( 5\,x-1 \right)  \left( 5-2\,x \right) }=18\,
x+32
$$
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