Can someone please explain in detail, how to find the limits of this inequality.
X^2 - | 2X - 1 | > 3X - 5
X^2 - | 2X - 1 | > 3X - 5
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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:seema19 wrote:Can someone please explain in detail, how to find the limits of this inequality.
X^2 - | 2X - 1 | > 3X - 5
Look at it this way: we start with x^2-|2x-1|>3x-5. Move everything to the left side: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?

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.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)
There is no need to 'complete the square' here, or to calculate discriminants - neither of those techniques are ever required in real GMAT questions. When x < 1/2, then |2x - 1| is equal to 1 - 2x. So our inequality becomesGmatMathPro wrote:
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.
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