Question

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

• A. $-\frac{1}{12}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{24}$

Answer 1

Answer: (A), (B)

$-\frac{1}{12}$
$\frac{1}{2}$

The given equation can be written as
$(x-\frac{1}{12})(x-\frac{1}{6})(x-\frac{1}{4})(x-\frac{1}{3})=\frac{5}{\mathrm{12.6.4.3}}$
...............(1)
Since $\frac{1}{12}<\frac{1}{6}<\frac{1}{4}<\frac{1}{3}$ and $\frac{1}{6}-\frac{1}{12}=\frac{1}{3}-\frac{1}{4}$,
we can introduce a new variable
$y=\frac{1}{4}[(x-\frac{1}{12})+(x-\frac{1}{6})+(x-\frac{1}{4})+(x-\frac{1}{3})]$
$y=x-\frac{5}{24}$
substituting $x=y+\frac{5}{24}$ in (1) we obtain
$(y+\frac{3}{24})(y+\frac{1}{24})(y-\frac{1}{24})(y-\frac{3}{24})=\frac{5}{\mathrm{12.6.4.3}}$
$\Rightarrow [y^2-{\left(\frac{1}{24}\right)}^2][y^2-{\left(\frac{3}{24}\right)}^2]=\frac{5}{\mathrm{12.6.4.3}}$
Hence we find that
$y^2=\frac{49}{{24}^2}$
i.e $y_1=\frac{7}{24}$ and $y_2=-\frac{7}{24}$
$\therefore$ The corresponding roots of the original equation are $-\frac{1}{12}$ and $\frac{1}{2}$