Is a solution of this equation a rational number?

Question

$$\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt[4]{4x-3}}=\frac{2}{3}$$

Are there any criteria of irreducibility of polynomials that handle this case?

Answer 1

First, let's move the first fraction across:

$$\frac{1}{\sqrt[4]{4x-3}} = \frac23-\frac{1}{\sqrt{2x-1}}$$

Now we can square both sides without making things worse:

$$\frac{1}{\sqrt{4x-3}} = \frac49 - \frac{4}{3\sqrt{2x-1}} + \frac{1}{2x-1}$$

$$\frac{1}{\sqrt{4x-3}} + \frac{4}{3\sqrt{2x-1}} = \frac{8x+5}{9(2x-1)}$$

Now we just have two square roots, so we square again:

$$\frac{1}{4x-3} + \frac{8}{3\sqrt{2x-1}\sqrt{4x-3}} + \frac{16}{9(2x-1)} = \frac{(8x+5)^2}{81(2x-1)^2}$$

$$\frac{8}{3\sqrt{2x-1}\sqrt{4x-3}} = \frac{64x^2 - 208x + 169}{81(2x-1)^2} - \frac{1}{4x-3}$$

$$\frac{8}{3\sqrt{(2x-1)(4x-3)}} = \frac{4 (64 x^3 - 337 x^2 + 406 x - 147)}{81(2x-1)^2(4x-3)}$$

$$\frac{2}{\sqrt{(2x-1)(4x-3)}} = \frac{64 x^3 - 337 x^2 + 406 x - 147}{27(2x-1)^2(4x-3)}$$

One more round of squaring:

$$\frac{4}{(2x-1)(4x-3)} = \frac{(64x^3 - 337x^2 + 406x - 147)^2}{729(2x-1)^4(4x-3)^2}$$

$$2916(2x-1)^3(4x-3) = (64x^3 - 337x^2 + 406x - 147)^2$$

Any root of the original equation is also a root of this polynomial, although the converse is not necessarily true. This simplifies to a lovely little sextic:

$$4096x^6 - 43136x^5 + 72225x^4 - 82508x^3 + 88954x^2 - 55212x + 12861 = 0$$

to which we can apply the rational root theorem. It has two real roots, which do not appear to be rational.

Answer 2

Let $y:=2x-1$. Note here that $x$ is rational if and only if $y$ is rational.

Since $4x-3=2y-1$, our equation becomes

$$\frac{1}{\sqrt{y}}+\frac{1}{\sqrt[4]{2y-1}}=\frac{2}{3}\tag1$$

We note here that both $Y=\frac{1}{\sqrt y}$ and $Y=\frac{1}{\sqrt[4]{2y-1}}$ are decreasing.

Supposing here that $y\ge 17$ gives $$\frac 23=\frac{1}{\sqrt{y}}+\frac{1}{\sqrt[4]{2y-1}}\le\frac{1}{\sqrt{17}}+\frac{1}{\sqrt[4]{2\cdot 17-1}}\lt 0.243+0.418\lt \frac 23$$ which is impossible where $$17\times (0.243)^2\gt 17\times 0.059=1.003\gt 1\implies \frac{1}{\sqrt{17}}\lt 0.243$$ $$33\times (0.418)^4\gt 33\times (0.1747)^2\gt 33\times 0.0305=1.0065\gt 1\implies \frac{1}{\sqrt[4]{2\cdot 17-1}}\lt 0.418$$

Also, supposing that $y\le 16$ gives $$\frac 23=\frac{1}{\sqrt{y}}+\frac{1}{\sqrt[4]{2y-1}}\ge\frac{1}{\sqrt{16}}+\frac{1}{\sqrt[4]{2\cdot 16-1}}\gt 0.25+0.42\gt\frac 23$$ which is impossible where $$31\times (0.42)^4\lt 31\times (0.177)^2\lt 31\times 0.032=0.992\lt 1\implies \frac{1}{\sqrt[4]{2\cdot 16-1}}\gt 0.42$$

So, we see that every real solution $y$ for $(1)$ satisfies $$16\lt y\lt 17\tag2$$

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Squaring the both sides of $$\frac{1}{\sqrt[4]{2y-1}}=\frac 23-\frac{1}{\sqrt y}$$ gives $$\frac{1}{\sqrt{2y-1}}=\frac{4y+9}{9y}-\frac{4}{3\sqrt y}$$ Squaring the both sides gives $$\frac{1}{2y-1}=\left(\frac{4y+9}{9y}\right)^2-\frac{8(4y+9)}{27y\sqrt y}+\frac{16}{9y},$$ i.e. $$\frac{8(4y+9)}{27y\sqrt y}=\left(\frac{4y+9}{9y}\right)^2+\frac{16}{9y}-\frac{1}{2y-1}=\frac{32y^3+335y^2-54y-81}{81y^2(2y-1)}$$ Multiplying the both sides by $81y^2(2y-1)$ gives $$24(4y+9)(2y-1)\sqrt y=32y^3+335y^2-54y-81$$ Squaring the both sides gives $$1024y^6-15424y^5-20255y^4-71316y^3+93798y^2-37908y+6561=0\tag3$$

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Since $1024=2^{10}$ and $6561=3^{8}$, by the rational root theorem, every positive rational solution for $(3)$ is of the form $$\frac{3^j}{2^i}\quad \text{$\quad $where $\ \ 0\le i\le 10,\ 0\le j\le 8$}$$

But none of these satisfies $(2)$ since $$\frac{3^3}{2^1}\lt16\lt 17\lt\frac{3^3}{2^0},\qquad \frac{3^4}{2^3}\lt16\lt 17\lt\frac{3^4}{2^2},\qquad \frac{3^5}{2^4}\lt16\lt 17\lt\frac{3^5}{2^3}$$ $$\frac{3^6}{2^6}\lt16\lt 17\lt\frac{3^6}{2^5},\qquad \frac{3^7}{2^8}\lt16\lt 17\lt\frac{3^7}{2^7},\qquad \frac{3^8}{2^9}\lt16\lt 17\lt\frac{3^8}{2^8}$$

So, $(1)$ has no rational solutions.

It follows from this that our equation has no rational solutions.

Answer 3

If we put $t=\sqrt[4]{4x-3}$ then we get $$4t^6-12t^5+9t^4-14t^2-12t+9=0$$

Since none of the candidates is rational root $t$ I assume that there is no rational root $x$ also.