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

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


   

Question 31166: Help!
Solve:
+sqrt%282x-1%29+%3C+x-2+
Thank you thank you thank you thank you thank you thank you thank (in advance!!)
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
+sqrt%282x-1%29+%3C+x-2+ then square both sides to give
+2x-1+%3C+%28x-2%29%5E2+
+2x-1+%3C+x%5E2-4x%2B4+
+0+%3C+x%5E2-6x%2B5+
or more usually re-ordered as +x%5E2-6x%2B5+%3E+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%28300%2C300%2C-2%2C8%2C-6%2C6%2Cx%5E2-6x%2B5%29

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.