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 $2a^2+5b^2=23$.
 A: 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 $x = 2.$
A: I will continue where you left off.
$$ a=\sqrt{5x−1} $$
We know that 
$$ \sqrt{5x−1} \geq 0$$
so
$$ x \geq  (1/5=0.2) ---- (Eq 1) $$
Given
$$2a^2+5b^2=23$$
Then
$$ 23-2a^2 \geq  0$$
Hence
$$ \frac{23}{2} \geq  a^2$$
That is:
$$\frac{23}{2} \geq  5x -1$$
Simplify to get:
$$x \leq 2.5$$
Combine this with (EQ 1) above, to get:
$$0.2 \leq x \leq 2.5$$
You may follow a similar process starting with $b$ to get a valid interval for $x$ and combing the two intervals you can get a close range of x. The problem can get simpler if you know that $x$ is an integer for example.
A: 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
$$ - \left( 26-x \right) \sqrt {5\,x-1}+ \left( 13\,x+14 \right) \sqrt {5
-2\,x}+12\,\sqrt { \left( 5\,x-1 \right)  \left( 5-2\,x \right) }=18\,
x+32
$$
A: 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?
