Question 31166
{{{ sqrt(2x-1) < x-2 }}} then square both sides to give
{{{ 2x-1 < (x-2)^2 }}}
{{{ 2x-1 < x^2-4x+4 }}}
{{{ 0 < x^2-6x+5 }}}
or more usually re-ordered as {{{ x^2-6x+5 > 0 }}}.


This is asking "which values of x in this quadratic give POSITIVE y-values" --> this is the meaning of the ">0"


First, lets find where the quadratic EQUALS zero

so (x-5)(x-1) = 0
so x-5=0 or x-1=0
so x=5 or x=1


So, we now know the position of the quadratic: {{{graph(300,300,-2,8,-6,6,x^2-6x+5)}}}


Looking at this, we can see that POSITIVE y-values (the thing asked for) happens when x is less than 1 or when x is greater than 5.


So answers are x<1 or x>5.


The only caveat i have is that the first line squares both sides: there may well be the possibility of multiplying by negative numbers here, in which case the inequality would swap round... i am hoping this is not required to be thought about for your answer.


jon.