SOLUTION: Help! Solve: {{{ sqrt(2x-1) < x-2 }}} Thank you thank you thank you thank you thank you thank you thank (in advance!!)

Question 31166: Help!
Solve:

Thank you thank you thank you thank you thank you thank you thank (in advance!!)
Answer by longjonsilver(2297) (Show Source): You can put this solution on YOUR website!
then square both sides to give

or more usually re-ordered as .

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:

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.

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