Positive roots of a polynomial

$(12x-1)(6x-1)(4x-1)(3x-1)=5$

$\therefore \left( (12x-1)(3x-1)\right)\left( (6x-1)(4x-1)\right)=5$

$\therefore (36x^{2}-15x+1)(24x^{2}-10x+1)=5$

$\therefore (3(12x^{2}-5x)+1)(2(12x^{2}-5x)+1)=5$

Let $12x^{2}-5x=t$

$\therefore (3t+1)(2t+1)=5$

$\therefore 6t^{2}+5t-4=0$

Solving the quadratic we obtain the roots as:

$t=\displaystyle\frac{1}{2}$ or $t=\displaystyle\frac{-4}{3}$

$\text{Case 1:}$

$12x^{2}-5x=\displaystyle\frac{1}{2}$

$\therefore 24x^{2}-10x+1=0$

$\therefore x=\displaystyle\frac{1}{2}$ or $x=\displaystyle\frac{-1}{12}$

Since only positive solutions are required we shall discard $x=\displaystyle\frac{-1}{12}$

$\text{Case 2:}$

$12x^{2}-5x=\displaystyle\frac{-4}{3}$

$\therefore 36x^{2}-15x+4=0$

As this equation yields complex roots we shall discard them.

Thus the required answer is $x=\boxed{0.5}$

I Have Very short method for This Question.

let

$12x-1=t$ .

so eqn becomes:

$t(t-1)(t-2)(t-3)=5*4*3*2$ .

$t=5$ .

$x=1/2 =0.5$ .

Q.E.D

$(12x-1)(6x-1)(4x-1)(3x-1)=5$

$\color{#ffffff}{\LaTeX}$

$\text{Multiply both sides by } 1\cdot 2 \cdot 3 \cdot 4$

$\displaystyle\color{#0000ff}{1\cdot}(12x-1)\color{#0000ff}{\cdot2\cdot}(6x-1)\color{#0000ff}{\cdot3\cdot}(4x-1)\color{#0000ff}{\cdot4\cdot}(3x-1)=5\color{#0000ff}{\cdot4\cdot}\color{#0000ff}{3\cdot}\color{#0000ff}{2\cdot}\color{#0000ff}{1}$

$(12x-1)(12x-2)(12x-3)(12x-4)=5\cdot4\cdot3\cdot2\cdot1$

$\color{#ffffff}{\LaTeX}$

$\text{Let }a=12x-1$

$(12x-1)(12x-1-1)(12x-1-2)(12x-1-3)=5\cdot4\cdot3\cdot2\cdot1$

$\color{#ffffff}{\LaTeX}$

$\text{Definition of a factorial: }$

$a(a-1)(a-2)(a-3)=5\cdot4\cdot3\cdot2\cdot1$

$a! = 5!$

$a = 5$

$5 = 12x - 1$

$\boxed{\color{#0000ff}{\displaystyle x=\frac{1}{2}}}$