SOLUTION: Hi {{{ (-(x-1)^2(2-x))/((3-x)(5x-1)^2)>=0 }}} It would be very nice to have explanations too. Thanks, Richard

Question 719218: Hi

It would be very nice to have explanations too.
Thanks,
Richard
Found 2 solutions by stanbon, lwsshak3:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
(-(x-1)^2(2-x))/((3-x)(5x-1)^2) >= 0
----
1st: Solve the EQUALITY:
-(x-1)^2(2-x) = 0
x = 1 or x = 2
---------
2nd: Note value that x cannot take:
x = 3 or x = 1/5
--------------------------
3rd: Solve the INEQUALITY:
Draw a number line and plot x = 1/5, x = 1, x = 2, x = 3
-----------------------------------
Test a value in each of the resulting 5 number line intervals
to find solution intervals:
(-(x-1)^2(2-x))/((3-x)(5x-1)^2) > 0
If x=0 you get: [-(-1)^2(2)]/[(3*(-1)^2] = -2/3 is not > 0
------
If x=1/2 you get: [-(-1/2)^2(3/2)]/[(5/2)(3/2)^2] = (-3/8)/+ is not > 0
------------
If x = 3/2 the result is not > 0
--------------------
If x = 5/2 the result IS > 0
--------
If x = 10 the result is not > 0
---
Conclusion:
Solution: x = 1 or values in the interval [2,3)
================================

========================
Cheers,
Stan H.
=========================

Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!

**
critical points:
x=1 (multiplicity 2)
x=2
x=3
x=1/5 (multiplicity 2)
Draw a number line with critical points on it:
<....-...1/5...-....1....-.....2...+.....3.....-.....>
Explanation:
When x>3, the expression becomes positive, but the negative sign in front makes it negative instead.
Going thru each critical point to the left, the intervals switch signs for odd multiplicities like 1,3, etc, but do not switch when going thru even multiplicities like 2,4, etc.
Solution: [2,3] (this is the only interval where x-values make the expression ≥0)
You could also do this with a chart but is is a much longer and tedious process.

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