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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote CRITICAL POINTS METHOD (For Inequalities) EXPLAINED- Useful! This topic has 8 member replies CRITICAL POINTS METHOD (For Inequalities) EXPLAINED- Useful! One often encounters expressions such as: (x - a)(x - b)(x - c)...(x -k) > 0 or < 0 in which we have to solve for x. Critical Points Method is the easiest way to solve such questions once you learn it. Let me teach you the method here. for the function f(x) = (x - a)(x - b)(x - c)...(x - k), Critical Points are those points at which f(x) = 0. So a, b, c.. k are the critical points of f(x). Example 1: Solve (x + 1).(x - 2) < 0 Step 1. Find out the critical points The critical points are -1 and 2 in this case. Step 2. Plot them on a number line in a proper order. Step 3: Consider the region above the number line to be positive and below it to be negative. Start drawing a curve starting from the right most - top side as shown below And then complete the curve in an alternate up-down fashion as shown below Step 4: Mark '+' in the region above the number line and '-' in the region below the number line as shown below. In the '+' region f(x) > 0 and in the '-' region f(x) < 0. Since the question wants us to solve for f(x) < 0, the values of x for which the curve goes below the number line will be chosen. The shaded region shown below is our answer. Answer: -1 < x < 2 Example 2: Lets take another question. (x - 3)(x - 2)(x - 1) > 0 Step 1: The critical points are 1, 2 and 3. Step 2: Mark them on a Number Line. Step 3: Plot the curve starting from top-right side in an alternate fashion. Step 4: Mark alternate '+' and '-' and choose the appropriate region In this case the region in which f(x) > 0 is favourable to us is shown below Answer: 1 < x < 2 , x > 3 Note: (i) If the expression is in the for (a - x)(x - b)(x - c) form, first convert it into (x - a)(x - b)(x - c) form by multiplying both sides of the inequality by -1 (and flipping the inequality sign). Apply the Critical Points method only after that. For example: (1 - 2x)(1 + x) < 0 has to be first converted into (2x - 1)(x + 1) > 0 form. (ii) If the inequality contains an 'equal to' sign as well such as (x - a)(x - b)(x - c) >=0 or =<0, we have to include the critical points in our region as a part of our solution. _________________ Aneesh Bangia GMAT Math Coach aneesh.bangia@gmail.com GMATPad: Facebook Page: https://www.facebook.com/GMATPad Master | Next Rank: 500 Posts Joined 06 Apr 2012 Posted: 134 messages Followed by: 5 members Upvotes: 35 aneesh.kg wrote: One often encounters expressions such as: (x - a)(x - b)(x - c)...(x -k) > 0 or < 0 in which we have to solve for x. Critical Points Method is the easiest way to solve such questions once you learn it. Let me teach you the method here. for the function f(x) = (x - a)(x - b)(x - c)...(x - k), Critical Points are those points at which f(x) = 0. So a, b, c.. k are the critical points of f(x). Example 1: Solve (x + 1).(x - 2) < 0 Step 1. Find out the critical points The critical points are -1 and 2 in this case. Step 2. Plot them on a number line in a proper order. Step 3: Consider the region above the number line to be positive and below it to be negative. Start drawing a curve starting from the right most - top side as shown below And then complete the curve in an alternate up-down fashion as shown below Step 4: Mark '+' in the region above the number line and '-' in the region below the number line as shown below. In the '+' region f(x) > 0 and in the '-' region f(x) < 0. Since the question wants us to solve for f(x) < 0, the values of x for which the curve goes below the number line will be chosen. The shaded region shown below is our answer. Answer: -1 < x < 2 Example 2: Lets take another question. (x - 3)(x - 2)(x - 1) > 0 Step 1: The critical points are 1, 2 and 3. Step 2: Mark them on a Number Line. Step 3: Plot the curve starting from top-right side in an alternate fashion. Step 4: Mark alternate '+' and '-' and choose the appropriate region In this case the region in which f(x) > 0 is favourable to us is shown below Answer: 1 < x < 2 , x > 3 Note: (i) If the expression is in the for (a - x)(x - b)(x - c) form, first convert it into (x - a)(x - b)(x - c) form by multiplying both sides of the inequality by -1 (and flipping the inequality sign). Apply the Critical Points method only after that. For example: (1 - 2x)(1 + x) < 0 has to be first converted into (2x - 1)(x + 1) > 0 form. (ii) If the inequality contains an 'equal to' sign as well such as (x - a)(x - b)(x - c) >=0 or =<0, we have to include the critical points in our region as a part of our solution. Thanks for sharing Aneesh! Useful1 Pl. have a look at my posts for 'Cheeky approach to few tricky problems' series 1-4 & 'Must see for Maximum/Minimum value problems'. Hope you like it. _________________ Shalabh Jain, e-GMAT Instructor Master | Next Rank: 500 Posts Joined 15 Feb 2011 Posted: 299 messages Followed by: 2 members Upvotes: 9 Please mention the scenarios when 1 --------------- > 0 (x-a)(x-b)(x-c) Master | Next Rank: 500 Posts Joined 16 Apr 2012 Posted: 385 messages Followed by: 29 members Upvotes: 186 If you notice, (x - 1)(x - 2)(x - 3) > 0 has been solved for x in Example 2 above. Since this post is generating some interest now, let me give a few practice problems to solve by this method: Solve for x in the following: i) x^2 > x ii) x^2 + 5x + 6 > 0 iii) (2 - x)(5x + 1)(x +2) < 0 I would like to see some attempts on these problems before giving away the solution. You can also validate your own answer by plugging-in numbers. _________________ Aneesh Bangia GMAT Math Coach aneesh.bangia@gmail.com GMATPad: Facebook Page: https://www.facebook.com/GMATPad Senior | Next Rank: 100 Posts Joined 09 Sep 2011 Posted: 32 messages Upvotes: 1 i) x^2 > x X>1 X<0 ii) x^2 + 5x + 6 > 0 = (x+3)(x+2) > 0 x>-2 x<-3 iii) (2 - x)(5x + 1)(x +2) < 0 =(x-2)(5x+1)(x+2) > 0 X > 2 -2 < X< -1/5 Your critical point technique is very useful, hopefully i got the answers correct, and my exam is coming up soon, and it seems this technique could come very handy! Senior | Next Rank: 100 Posts Joined 09 Sep 2011 Posted: 32 messages Upvotes: 1 Quick question, do you always start drawing the line coming from the top right, could something like this ever exist?: Master | Next Rank: 500 Posts Joined 13 Jun 2011 Posted: 114 messages Upvotes: 1 Dear Aneesh Thanks very much. This is a very good trick. Can we extend this trick for two variables i.e, (x+1)(y+2)>0 with your trick the soln will be x > -1 and y < -2 Master | Next Rank: 500 Posts Joined 16 Apr 2012 Posted: 385 messages Followed by: 29 members Upvotes: 186 Hi zueswoods, First of all, All the best for your GMAT! Good question, and I'm glad you asked. I asked you to convert everything into a (x - a)(x - b)(x - c)... form if it is in (a - x)(x - b)(x - c).. form so that you don't have to worry about the possibility of the curve starting from top bottom. Once you've done the conversion (if required, by multiplying by a -1) then the curve will always start with a positive on the right side. If the inequality is (a - x)(b - x) < 0 then you've to multiply it twice by -1 to reduce it to: (x - a)(b - x)< 0. Yeah, you get the same thing. But, do it anyway so that you stay in the habit of bringing it to the required form. Cheers! _________________ Aneesh Bangia GMAT Math Coach aneesh.bangia@gmail.com GMATPad: Facebook Page: https://www.facebook.com/GMATPad Master | Next Rank: 500 Posts Joined 16 Apr 2012 Posted: 385 messages Followed by: 29 members Upvotes: 186 Hi Krishna, If you apply it to two variables, you will not get a complete solution so Avoid it! Let's take the example presented by you: (x + 1)(y + 2) > 0 This inequality holds true in these cases: Case1: x + 1 > 0, or x > -1 and y + 2 > 0, or y > -2 Case2: x + 1 < 0, or x < -1 and y + 2 < 0, or y < -2 So, if you apply Critical Points to this inequality, you will miss out Case2. Moral of the Story: Apply this method for one variable only. _________________ Aneesh Bangia GMAT Math Coach aneesh.bangia@gmail.com GMATPad: Facebook Page: https://www.facebook.com/GMATPad • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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