Problem Solving

Can someone please explain in detail, how to find the limits of this inequality.

X^2 - | 2X - 1 | > 3X - 5

three cases are considered
{2x-1>0
{2x-1=0
{2x-1<0
Case 2x-1>0
x^2-2x+1>3x-5, (x-3)(x-2)>0
Cases 2x-1=0 and 2x-1<0 have no critical points hence are not considered.

Only Case 2x-1>0 is considered at this time.

(x-3)(x-2)>0
Possible
a) x-3>0 and x-2>0, (x>3 U x>2) --> x>3 is one limit
b) x-3<0 and x-2<0, (x<3 U x<2) --> x<2 is another limit

the patterns a), b) should be tested by plugging the values for x into inequality to find the relevant interval(s). This inequality would have solution area (-infinity;2)U(3;+infinity), but because 2x-1>0 and our mode doesn't have critical points when 2x-1=0 and 2x-1<0 we exclude x=<1/2 from the solution area, hence our answer (limits) would be (1/2;2)U(3;+infinity)

graphically would be ___-infinity_xxxxxx_0_xxxxxx_1/2_____2_xxxxxx_3_________+infinity___
the areas xxxxx are not solution areas.

seema19 wrote:Can someone please explain in detail, how to find the limits of this inequality.

X^2 - | 2X - 1 | > 3X - 5

seema19 wrote:Can someone please explain in detail, how to find the limits of this inequality.

X^2 - | 2X - 1 | > 3X - 5

You need to rewrite the inequality without the absolute value bars by considering different cases:

Case #1: 2x-1>=0 or x>=1/2.

If we ONLY consider values of x such that 2x-1 is greater than or equal to 0, then the absolute value bars have no effect on the expression 2x-1 because it is already definitely not negative, thus, they may be dropped. For values of x>=1/2, the above expression is equivalent to: x^2-(2x-1)>3x-5. Solve this as follows:

x^2-2x+1>3x-5

x^2-5x+6>0 (bring everything to one side)

(x-3)(x-2)>0 (factor)

this expression is true if x<2 or x>3. Because the expression we solved was equivalent to the original expression only if x>=1/2, then we have to include this restriction in our solution set. This gives 1/2<=x<2 OR x>3.

Case #2: 2x-1<0 or x<1/2.

Notice that if we only consider values of x where x<1/2, then 2x-1 is ALWAYS negative. Taking the absolute value of a negative number is the same as multiplying that number by -1. For example |-3| is the same as -1*-3, so |2x-1| is the same as -1*(2x-1), which equals 1-2x if we can be sure 2x-1 is negative. Thus, we can rewrite the original expression as x^2-(1-2x)>3x-5. Solving:

x^2-1+2x>3x-5

x^2-x+4>0

By completing the square, we can rewrite x^2-x+4 as (x-1/2)^2 + 3.75, so we have:

(x-1/2)^2+3.75>0.

This is always true because a squared quantity plus a positive number is always positive. We arrived at this expression by only considering values of x such that x<1/2. Thus, this equivalent inequality tells us that all x<1/2 should be part of our solution set.

Summarizing, the first case tells us that values of x such that 1/2<=x<2 or x>3 are part of the solution. The second case tells us that all values less than 1/2 are solutions. Thus, the final overall solution to the original inequality will be the combination of the intervals we got from the two cases:

Ans: [spoiler]x<2 or x>3[/spoiler]

@Pete, I thought we are restricted by mode signs on the interval x<=1/2 and I've joint two intervals (1/2;2)U(3;+infinity). Don't we have no critical values for the values to the left of 1/2 on the coordinate line (because x-2<0 and x-2=0, the discriminants in their quadratics set to find the critical values are -ve), and when we consider the original inequality X^2 - | 2X - 1 | > 3X - 5 we have to set not only solution area for x-2>0 but for all cases?

pemdas wrote:@Pete, I thought we are restricted by mode signs on the interval x<=1/2 and I've joint two intervals (1/2;2)U(3;+infinity). Don't we have no critical values for the values to the left of 1/2 on the coordinate line (because x-2<0 and x-2=0, the discriminants in their quadratics set to find the critical values are -ve), and when we consider the original inequality X^2 - | 2X - 1 | > 3X - 5 we have to set not only solution area for x-2>0 but for all cases?

Look at it this way: we start with x^2-|2x-1|>3x-5. Move everything to the left side:

x^2-|2x-1|-3x+5>0. Now, consider the graph of y=x^2-|2x-1|-3x+5. I made the following graph at https://www.graph.tk if you want to reproduce it:



Because y=x^2-|2x-1|-3x+5 and our inequality is x^2-|2x-1|-3x+5>0, the original inequality will be true when y>0. From the graph it is clear that this is true whenever x<2 or x>3. There's no reason to exclude the solutions that are less than 1/2. What's more, you can verify that there are solutions that are less than 1/2 by plugging in values:

x^2-|2x-1|>3x-5

x=0: -1>-5

x=-1: -2>-8

x=-9.5: 70.25>-33.5

All of them are solutions to the original inequality and must be included in any solution set.

clear, i see
strangely, i caught myself on changing question answers marked towards the end and before moving to other questions, like here and in many of my mock tests. Mostly, I was able to resolve and get the right answers but then was marking different choices. Sometimes adding unnecessary restrictions to combs/perms like on one forum you contributed greatly, or arguing about irrelevant properties, I need more efficient approach. Any advice you could give on my occasions? (I'm asking in this forum, because the issue is specific and mostly quant related)

pemdas wrote:clear, i see
strangely, i caught myself on changing question answers marked towards the end and before moving to other questions, like here and in many of my mock tests. Mostly, I was able to resolve and get the right answers but then was marking different choices. Sometimes adding unnecessary restrictions to combs/perms like on one forum you contributed greatly, or arguing about irrelevant properties, I need more efficient approach. Any advice you could give on my occasions? (I'm asking in this forum, because the issue is specific and mostly quant related)

So you're saying on tests you'll solve a problem, mark an answer, and then before hitting submit, you solve it again? And then you usually end up changing it so that it's wrong? Or do you usually fix it so it's right? I was a little unclear on this point.

If you find yourself going back and changing your answers a lot, it may just be a matter of building up your confidence in certain areas. From your posts it sounds like you have a pretty strong math background, particularly in statistics. Is this correct? But maybe in some areas there's still room to improve your fundamentals.

I think permutations and combinations is just hard for almost everybody, and the vast majority of people wouldn't feel that confident in a purely theoretical approach to hard counting problems until they've spent years working with them. Obviously, most people don't have the luxury of doing that before taking the GMAT, so I think the next best thing is to have a backup plan for those types of problem. That is, don't rely on a theoretical approach alone. Take your best guess as to what the correct method is, and then test that method on a similar, more accessible problem. For example, in that 6 men, 4 women around a table problem, I listed out the possibilities for 4 men, 2 women to demonstrate that my method worked. Obviously you can't list out 72 possibilities on a real test, but you could list out all the possibilities for 3 men and 2 women, as there are only 12 possibilities. If the method you were using also returned 12 for that case, that would give you confidence that you're on the right track. Also, examining a complete list of possibilities for a specific instance like this could give you a lot of information about whether assumptions that you are making are reasonable or unreasonable.

In the case of this inequality, it looks like you again got in trouble by relying on an overly theoretical approach. You were correct that the discriminant for the quadratic that resulted for cases where x<1/2 was negative, however, you applied this information incorrectly. The fact that the discriminant is negative just means that the quadratic expression can never equal zero, so it is in fact either always positive or always negative. In any case, it is not difficult to plug in a couple of values less than 1/2 to see if the results support your theoretical approach. If the test fails, you have an opportunity to reexamine your assumptions.

In sum, try to get in the habit of finding ways to test your theoretical approaches to the harder problems by examining more accessible situations or by doing things like plugging in numbers. It may take a little longer, but it should increase your accuracy and confidence in your solution.

X^2 - | 2X - 1 | > 3X - 5

You should do such ques by considering cases on the number line

The expression |2x-1| can give yoou two cases
case 1:
2x - 1 < 0
x < 1/2

If 2x-1 is negative

|2x-1| = -(2x-1)

Using this in the inequation
x^2 - (-(2x-1) > 3x-5
x^2 + 2x - 1 > 3x-5
x^2 - x + 4 > 0

The dscriminant is negative
which means it satisfies all values frm -ve infinity to +ve infinity
this is for x<1/2
so the limits in this case would be
x < 1/2

Case 2:
2x - 1 >= 0
x >= 1/2

x^2 - 2x +1 > 3x - 5
x^2 -5x +6 > 0
(x-2)(x-3) > 0
x >= 1/2

from above inequalities
1/2 <= x < 2 and 3< x


Combining both the cases we get

x < 2 U x > 3

Good question