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Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
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Re: Inequalities trick [#permalink]
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02 Jun 2017, 08:46
Shiv2016 wrote: Hi karishma, abhimahna, mikemcgarry, Bunuel, Skywalker18, and other experts I am not able to understand the graph. How do we know which one is negative/positive? How does this graph help us in solving inequalities? Thanks The links of the three posts given here should clear all your doubts: https://gmatclub.com/forum/inequalities ... l#p1753431
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Re: Inequalities trick [#permalink]
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02 Jun 2017, 09:16
abhimahna wrote: Shiv2016 wrote: Hi karishma, abhimahna, mikemcgarry, Bunuel, Skywalker18, and other experts I am not able to understand the graph. How do we know which one is negative/positive? How does this graph help us in solving inequalities? Thanks Hi Shiv2016 , The concept is very easy. 1. Find out the zero points. 2. Arrange them in ascending order. 3. Draw them on a number line. 4. Take the right most as +ve and proceed towards left taking alternate signs. 5. If the inequality is of > form, then take all +ve ranges. 6. if the inequality is of lesser form, then take all ve ranges. E.g. (x2) (x3 ) > 0 Equality form is greater than(>). Zero points = 2 and 3 Draw them on number line. You will get 3 ranges. x<2; 2<x<3; and x >2. Here, right most will be +ve, or x>2 will be +ve. then 2<x<3 will be ve then x<2 will be +ve. Since inequality is of (>) form, we will take all the ranges which have +ve sign. Hence, the answer will be x>2 and x<2. I hope it makes sense. Thanks for your reply but I think I am still there.. If (x2)(x3)>0 Then won't the answer be: x>2 and x>3 So we will take x>3 Why is this wrong as we have been solving questions this way? I am sorry but there is some gap in my understanding (i THINK)



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Re: Inequalities trick [#permalink]
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02 Jun 2017, 09:32
Shiv2016 wrote: Hi
I am not able to understand the graph. How do we know which one is negative/positive? How does this graph help us in solving inequalities?
Thanks Dear Shiv2016, I'm happy to respond. I see that abhimahna already gave I wonderful explanation. I will just add a few points. The post at the beginning of the thread is not 100% accurate and it is way to general for the GMAT. Inequalities with powers of x is a exceeding rare topic on the GMAT, and when there is a power in an inequality, it almost always is just x. I think you could take 50 GMATs in a row and not see a power of x higher than 2 in an inequality. You may find this blog helpful: GMAT Quadratic InequalitiesI definitely agree that finding the roots, the xvalues that make the function zero, is the first step. Find these, and put these on a number line. If the highest power is simply 2, then you should know that graph is a parabola. If the coefficient of \(x^2\) is positive, then the parabola opens upward, like the letter U. If the coefficient of \(x^2\) is negative, then the parabola opens downward, like an upsidedown U. Right there, that should tell you where the function is positive or negative. Higher powers of x (\(x^3\), \(x^4\), etc.) appear infrequently. The alternating pattern doesn't always work: in particular, the signs don't alternate if there's a repeated root, e.g. \(x^3  4x^2 + 4x > 0\)). That's a caveat for higher mathematics, but I don't think anyone would ever need to know this for the GMAT. My friend, part of the reason you were confused is because the person who opened the thread was talking about more advanced math than you need for the GMAT. People who study engineering learn way more math than is needed for the GMAT, and some of them like to show off, but this doesn't really help anyone. Does all this make sense? Mike
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Re: Inequalities trick [#permalink]
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02 Jun 2017, 09:44
mikemcgarry wrote: Shiv2016 wrote: Hi
I am not able to understand the graph. How do we know which one is negative/positive? How does this graph help us in solving inequalities?
Thanks Dear Shiv2016, I'm happy to respond. I see that abhimahna already gave I wonderful explanation. I will just add a few points. The post at the beginning of the thread is not 100% accurate and it is way to general for the GMAT. Inequalities with powers of x is a exceeding rare topic on the GMAT, and when there is a power in an inequality, it almost always is just x. I think you could take 50 GMATs in a row and not see a power of x higher than 2 in an inequality. You may find this blog helpful: GMAT Quadratic InequalitiesI definitely agree that finding the roots, the xvalues that make the function zero, is the first step. Find these, and put these on a number line. If the highest power is simply 2, then you should know that graph is a parabola. If the coefficient of \(x^2\) is positive, then the parabola opens upward, like the letter U. If the coefficient of \(x^2\) is negative, then the parabola opens downward, like an upsidedown U. Right there, that should tell you where the function is positive or negative. Higher powers of x (\(x^3\), \(x^4\), etc.) appear infrequently. The alternating pattern doesn't always work: in particular, the signs don't alternate if there's a repeated root, e.g. \(x^3  4x^2 + 4x > 0\)). That's a caveat for higher mathematics, but I don't think anyone would ever need to know this for the GMAT. My friend, part of the reason you were confused is because the person who opened the thread was talking about more advanced math than you need for the GMAT. People who study engineering learn way more math than is needed for the GMAT, and some of them like to show off, but this doesn't really help anyone. Does all this make sense? Mike Thanks for your reply Mike. Yes the first post (though might be helpful) but really makes the quant tougher to understand. I actually came across this post because I saw the same graph in egmat file and was not able to understand it. I know few very basic facts: 1) On a number line, zero is the mid point and nos. on left of zero are negative and on right of zero are positive. 2) In an inequality, we have to solve for values of x in order to find its range. But this graph kind of contradicts with point 1 and that is what confused me.



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Re: Inequalities trick [#permalink]
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02 Jun 2017, 11:52
Shiv2016 wrote: I know few very basic facts: 1) On a number line, zero is the mid point and nos. on left of zero are negative and on right of zero are positive. 2) In an inequality, we have to solve for values of x in order to find its range. But this graph kind of contradicts with point 1 and that is what confused me. No it does not. These are apples and oranges. Zero Is and Will Always BE the midpoint on the number line. However, we do not care for Zero in these questions unless Zero is one of the solutions of the equation. Let me take the following example: Shiv2016 wrote: If (x2)(x3)>0
Then won't the answer be: x>2 and x>3
So we will take x>3
Why is this wrong as we have been solving questions this way? I am sorry but there is some gap in my understanding (i THINK) If the equation were (x2)(x3) = 0 Then, x = 2 and 3 Answer But the Equation is (x2)(x3) > 0 So, the sign of the will switch in following way: +++++++++ 2  3 ++++++++++ (Number Line) We did not mark Zero as it is not one of the solutions of the equation.Therefore, for all values of x < 2 And x > 3, inequality (x2)(x3) > 0 will be true.
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Re: Inequalities trick [#permalink]
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09 Jun 2017, 03:44
I found this as well as the following post very helpful! https://gmatclub.com/forum/inequations ... 54664.htmlCan someone explain how this works with absolute value inequalities?



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Re: Inequalities trick [#permalink]
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01 Dec 2017, 02:08
VeritasPrepKarishma wrote: Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain. If (xa)(xb)(xc)(xd) < 0, we can draw the points a, b, c and d on the number line. e.g. Given (x+2)(x1)(x7)(x4) < 0, draw the points 2, 1, 7 and 4 on the number line as shown. Attachment: doc.jpg This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive. When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x  7) will be negative since x < 7 in this region. All other factors are still positive. When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x  7) and (x  4) are negative, but negative x negative is positive... and so on till you reach the leftmost section. Since we are looking for values of x where the expression is < 0, here the solution will be 2 < x < 1 or 4< x < 7 It should be obvious that it will also work in cases where factors are divided. e.g. (x  a)(x  b)/(x  c)(x  d) < 0 (x + 2)(x  1)/(x 4)(x  7) < 0 will have exactly the same solution as above. Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion. Responding to a pm: Quote: hii can u plz elaborate what happens when we have <= or >=, in division as you said the solution will differ.
Note that in case of division, the denominator cannot be 0. So given \(\frac{(x+2)(x1)}{(x7)(x4)} <= 0\) We will use the same method to find that the solution will be 2 < x < 1 or 4< x < 7 but here, we have an equal to sign too. So the expression can be equal to 0 too. So x can be equal to 2 and it can be equal to 1 too. But x cannot be 4 and neither can it be 7 since that will make the denominator 0. So the actual solution here will be 2 <= x <= 1 or 4< x < 7
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