Question 1049350
Use the definition of the absolute value function to break up the function into two functions.
1.{{{x(5x-1)=2x-4}}} when {{{5x-1>=0}}}
2.{{{-x(5x-1)=2x-4}}} when {{{5x-1<0}}}
So then,
Using eq. 1,
{{{5x^2-x=2x-4}}}
{{{5x^2-3x+4=0}}}
No real solutions.
Using eq. 2,
{{{-5x^2+x=2x-4}}}
{{{-5x^2-x+4=0}}}
{{{5x^2+x-4=0}}}
{{{(x+1)(5x-4)=0}}}
Two solutions:
{{{x+1=0}}}
{{{x=-1}}}
and
{{{5x-4=0}}}
{{{5x=4}}}
{{{x=4/5}}}
However we have to check our original condition,
{{{5x-1<0}}}
{{{5x<1}}}
{{{x<1/5}}}
So then only {{{x=-1}}} would be a solution.