Question

The value(s) of $y$ satisfying the equation
$|\frac{y-4}{y^2}|=1\text{is/are}:$

Answer 1

Given equation:
$|\frac{y-4}{y^2}|=1$
$\Rightarrow \frac{|y-4|}{|y^2|}=1\Rightarrow |y^2|=|y-4|\Rightarrow |y^2|-|y-4|=0$
We know that
$\because |x^2|=x^2$
So
$\Rightarrow y^2-|y-4|=0$
Case $1:\text{When}y\ge 4$
$\Rightarrow y^2=y-4$
$\Rightarrow y^2-y+4=0$
For this quadratic equation
$D=-15<0$
Thus, no real solution exists for this case.
Case $2:\text{When}y<4$
$\Rightarrow y^2=-(y-4)$
$\Rightarrow y^2+y-4=0$
For this quadratic equation,
$D=17>0$
So, real solutions exists for this case.
We know that for quadratic equation
$a{y}^2+by+c=0;\text{where}a\ne 0$
$y=\frac{-b\pm \sqrt{D}}{2a}$
Thus, roots of the equation will be:
$y=\frac{-1\pm \sqrt{17}}{2}$