Radical equation $\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt[4]{4x-3}}=\frac{2}{3}$ I am stuck in the following difficult radical equation:
$$
\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt[4]{4x-3}}=\frac{2}{3}
$$
Attemption. The left hand side of the equation is a decrease function. Therefore, the equation has a unique solution. An approximation for the solution is 8.74874 (using Matlab).
Thank you for all kind help hint.
 A: As Donald Splutterwit points out it his comment, the problem reduces to a polynomial equation of the sixth degree. Generally, there are no formulaic solutions to polynomial equations of the fifth and higher degrees. However, even where there is a formula, for example in the case of a quadratic equation such as $x^2=2$, actually writing down a solution as a number in the decimal system requires approximative methods, such as Newton–Raphson (NR).
The problem you state is very convenient to solve by NR, practically in its original form, to the accuracy that one's computational facilities allow. Thus, the task is to find a zero of the function $$f(x):=(2x-1)^{-1/2}+(4x-3)^{-1/4}-\tfrac23$$by NR. This is probably what the Matlab program does (perhaps with some streamlining shortcuts), and a human being cannot do better.
A: Hint: $u=2x-1$ and $v=4x-3$ 
Rewrite the expression:
$$
\frac{1}{\sqrt{u}}+\frac{1}{\sqrt[4]{v}}=\frac{2}{3} \implies \frac{\sqrt[4]{v}+\sqrt{u}}{\sqrt{u}\sqrt[4]{v}}=\frac{2}{3}
$$
$$
\implies \sqrt[4]{v}+\sqrt{u}=\frac 23\sqrt{u}\sqrt[4]{v}
$$
$$
\implies \sqrt{u}=\left(\frac 23\sqrt{u}-1\right)\sqrt[4]{v} \implies \frac{\sqrt{u}}{\frac 23\sqrt{u}-1}=\sqrt[4]{v} 
$$
Now, raise the equation to the fourth power:
$$
\implies \frac{u^2}{(\frac 23\sqrt{u}-1)^4}=v 
$$
Now isolate $\sqrt{u}$ on one side of the equation and raise everything to the second power. The resulting polynomial in $x$ after inverting the substitutions might not be solvable by an analytical expression.
A: Making the substitution: $4x-3 = t^4$ , then : $2x-1= \frac{t^4+1}2$.
After doing a lot of algebraic transformations,(squaring both sides two times) 
I have obtained the $12$-th grade polynomial:
$$16 t^{12}- 72 t^{10} - 31 t^8- 468 t^6 + 358 t^4 - 396 t^2 + 81 = 0.$$
Solving via Mathematica FindRoot function, 
the only real roots that give the correct result are $t=\pm 2.37832$
Back substitution give us: $x=8.748732273$
