SOLUTION: Please tell the way to solve this : x^2-2x > |2x-1|. Thank you

Algebra ->  Radicals -> SOLUTION: Please tell the way to solve this : x^2-2x > |2x-1|. Thank you      Log On


   

Question 989123: Please tell the way to solve this : x^2-2x > |2x-1|. Thank you
Found 2 solutions by josgarithmetic, solver91311:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
The only way is to use the two possible conditions for 2x-1 and solve each resulting inequality.

2x-1 NON-NEGATIVE
x%5E2-2x%3E2x-1
x%5E2-4x%2B1%3E0
Parabola concave up, inequality will be false between the roots.

2x-1 NEGATIVE
x%5E2-2x%3E-2x%2B1
x%5E2%3E1
x%5E2-1%3E0
Inequality will be false between the roots.

These solutions may be combined with OR.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


To keep yourself from getting all balled up with absolute value bars, intersections and unions of solution sets, and sundry other bumps in this particular road, the best thing to do is to get a visual idea of what the answer is going to look like before you start working out the details. Hence, graph the two sides of your inequality on the same set of coordinate axes.

.

The magenta line is the graph of the quadratic function and the red line is the graph of the absolute value function.

We want to find the intervals where the quadratic function is larger than the absolute value function.



Recalling the definition of absolute value, we can create two inequalties:



And



Let's begin with the second one and add 2x to both sides



Noting that the zeros of are , we can see that must either be less than -1 or greater than 1. But if you look back at the graph, you can see that 1 is an extraneous root since nowhere in the vicinity of 1 is the quadratic function larger than the absolute value function. Hence, for THIS HALF of the problem, the appropriate interval is . Note that the interval is open on both ends because the original inequality is strictly greater than.

Looking at the other half of the problem:





I'll leave it as an exercise for you to verify that the zeros of the corresponding quadratic equation are .

Again, we have an extraneous root. is a number between 0 and 1 where the quadratic function is everywhere less than the absolute value function. However, is the point where the two functions intersect and after which the quadratic begins to be the larger valued function. Hence, the appropriate interval for this half of the problem is . Again, the "strictly greater than" relation in the original inequality demands an open interval.

Taken together the solution set is the union of the two intervals:



John

My calculator said it, I believe it, that settles it