Question

Find $x$.
$3x+\frac{5}{16x}-2=0$

Answer 1

$3x+\frac{5}{16x}-2=0$

$\frac{5}{16x}=2-3x$

$5=\left(16x\right)(2-3x)$

$\Rightarrow$ $5=32x-48x^2$

$\Rightarrow$ $48x^2-32+5=0$

$\Rightarrow$ $48x^2-12x-20x+5=0$

$\Rightarrow$ $12x(4x-1)-5(4x-1)=0$

$\Rightarrow$ $(4x-1)(12x-5)=0$

$\Rightarrow$ $4x-1=0$ or $1‘2x-5=0$

$\Rightarrow$ $x=\frac{1}{4}$ or $x=\frac{5}{12}$

$\therefore$ roots of $3x+\frac{5}{16x}-2=0$ are $(\frac{1}{4},\frac{5}{12})$