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If x is positive, is x > 3
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If x is positive, is x > 3 ? (1) (x  1)^2 > 4 (2) (x  2)^2 > 9 Can someone point a mistake in my method? (1) Taking one of the equations: (x  1)^2 > 4 x^2 + 1  2x > 4 x^2 + 1  2x  4 > 0 x^2  3x + 1x  3 > 0 (x3) (x+1) > 0 x > 3 and x > 1
while in the explanation given the ans is coming out to be x > 3 and x < 1
Please help ..
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Originally posted by rohitgoel15 on 16 Apr 2012, 09:42.
Last edited by ENGRTOMBA2018 on 14 Oct 2015, 16:14, edited 1 time in total.
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If x is positive, is x > 3
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16 Apr 2012, 11:58
rohitgoel15 wrote: If x is positive, is x > 3 ?
(1) (x  1)^2 > 4 (2) (x  2)^2 > 9
Can someone point a mistake in my method? (1) Taking one of the equations: (x  1)^2 > 4 x^2 + 1  2x > 4 x^2 + 1  2x  4 > 0 x^2  3x + 1x  3 > 0 (x3) (x+1) > 0 x > 3 and x > 1
while in the explanation given the ans is coming out to be x > 3 and x < 1
Please help .. If x is positive, is x > 3 ?(1) (x  1)^2 > 4 > \((x+1)(x3)>0\) > roots are 1 and 3. Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<1\) or \(x>3\). Since given that \(x\) is positive then only one range is valid: \(x>3\). Sufficient. (2) (x  2)^2 > 9 > \((x+1)(x5)>0\) > roots are 1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<1\) or \(x>5\). Since given that \(x\) is positive then only one range is valid: \(x>5\). Sufficient. Answer: D. Solving inequalities: http://gmatclub.com/forum/x24x94661.html#p731476 ( check this one first) http://gmatclub.com/forum/inequalitiestrick91482.htmlhttp://gmatclub.com/forum/datasuffine ... 09078.htmlhttp://gmatclub.com/forum/rangeforvar ... me#p873535http://gmatclub.com/forum/everythingis ... me#p868863Another approach:If x is positive, is x > 3 ?(1) (x  1)^2 > 4 > since both sides of the inequality are nonnegative then we can take square root from both parts: \(x1>2\). \(x1\) is just the distance between 1 and \(x\) on the number line. We are told that this distance is more than 2: (1)13 so, \(x<1\) or \(x>3\). Since given that \(x\) is positive then only one range is valid: \(x>3\). Sufficient. (2) (x  2)^2 > 9 > \(x2>3\). The same here: \(x2\) is just the distance between 2 and \(x\) on the number line. We are told that this distance is more than 3: (1)25 so, \(x<1\) or \(x>5\). Since given that \(x\) is positive then only one range is valid: \(x>5\). Sufficient. Answer: D. Hope it helps.
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Re: If x is positive, is x > 3
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20 Jul 2013, 06:57
Hi bunuel, Just wanted to clarify in the alternative approach you mentioned "nonnegative" so if the other side of the inequality has a negative number, the only way to proceed with the problem is by expansion?
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Re: If x is positive, is x > 3
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20 Jul 2013, 07:32
fozzzy wrote: Hi bunuel,
Just wanted to clarify in the alternative approach you mentioned "nonnegative" so if the other side of the inequality has a negative number, the only way to proceed with the problem is by expansion? If it were (x  1)^2 > 4, it would simply mean that x can take any value. As for general rules for inequalities: taking the square root, squaring, ... ADDING/SUBTRACTING INEQUALITIES: You can only add inequalities when their signs are in the same direction:If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) > \(a+c>b+d\). Example: \(3<4\) and \(2<5\) > \(3+2<4+5\). You can only apply subtraction when their signs are in the opposite directions:If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). RAISING INEQUALITIES TO EVEN/ODD POWER: A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are nonnegative (the same for taking an even root of both sides of an inequality).For example: \(2<4\) > we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) > we can square both sides and write: \(x^2<y^2\); But if either of side is negative then raising to even power doesn't always work. For example: \(1>2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are nonnegative. B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).For example: \(2<1\) > we can raise both sides to third power and write: \(2^3=8<1=1^3\) or \(5<1\) > \(5^2=125<1=1^3\); \(x<y\) > we can raise both sides to third power and write: \(x^3<y^3\). For multiplication check here: helpwithaddsubtractmultdividmultipleinequalities155290.html#p1242652THEORY ON INEQUALITIES: x24x94661.html#p731476inequalitiestrick91482.htmldatasuffinequalities109078.htmlrangeforvariablexinagiveninequality109468.htmleverythingislessthanzero108884.htmlgraphicapproachtoproblemswithinequalities68037.htmlinequationsinequalitiespart154664.htmlinequationsinequalitiespart154738.htmlQUESTIONS: All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189700+ Inequalities problems: inequalityandabsolutevaluequestionsfrommycollection86939.htmlHope it helps.
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Re: If x is positive, is x > 3
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05 Sep 2013, 03:47
Bunuel wrote: rohitgoel15 wrote: If x is positive, is x > 3 ?
(1) (x  1)^2 > 4 (2) (x  2)^2 > 9
Can someone point a mistake in my method? (1) Taking one of the equations: (x  1)^2 > 4 x^2 + 1  2x > 4 x^2 + 1  2x  4 > 0 x^2  3x + 1x  3 > 0 (x3) (x+1) > 0 x > 3 and x > 1
while in the explanation given the ans is coming out to be x > 3 and x < 1
Please help .. If x is positive, is x > 3 ?(1) (x  1)^2 > 4 > \((x+1)(x3)>0\) > roots are 1 and 3. Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<1\) or \(x>3\). Since given that \(x\) is positive then only one range is valid: \(x>3\). Sufficient. (2) (x  2)^2 > 9 > \((x+1)(x5)>0\) > roots are 1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<1\) or \(x>5\). Since given that \(x\) is positive then only one range is valid: \(x>5\). Sufficient. Answer: D. Solving inequalities: x24x94661.html#p731476 ( check this one first) inequalitiestrick91482.htmldatasuffinequalities109078.htmlrangeforvariablexinagiveninequality109468.html?hilit=extreme#p873535everythingislessthanzero108884.html?hilit=extreme#p868863Hope it helps. Another approach:If x is positive, is x > 3 ?(1) (x  1)^2 > 4 > since both sides of the inequality are nonnegative then we can take square root from both parts: \(x1>2\). \(x1\) is just the distance between 1 and \(x\) on the number line. We are told that this distance is more than 2: (1)13 so, \(x<1\) or \(x>3\). Since given that \(x\) is positive then only one range is valid: \(x>3\). Sufficient. (2) (x  2)^2 > 9 > \(x2>3\). The same here: \(x2\) is just the distance between 2 and \(x\) on the number line. We are told that this distance is more than 3: (1)25 so, \(x<1\) or \(x>5\). Since given that \(x\) is positive then only one range is valid: \(x>5\). Sufficient. Answer: D. I used x= +6 and 6 ..Which is true in both the cases..it shud be E..



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Re: If x is positive, is x > 3
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05 Sep 2013, 03:52
SUNGMAT710 wrote: Bunuel wrote: rohitgoel15 wrote: If x is positive, is x > 3 ?
(1) (x  1)^2 > 4 (2) (x  2)^2 > 9
Can someone point a mistake in my method? (1) Taking one of the equations: (x  1)^2 > 4 x^2 + 1  2x > 4 x^2 + 1  2x  4 > 0 x^2  3x + 1x  3 > 0 (x3) (x+1) > 0 x > 3 and x > 1
while in the explanation given the ans is coming out to be x > 3 and x < 1
Please help .. If x is positive, is x > 3 ?(1) (x  1)^2 > 4 > \((x+1)(x3)>0\) > roots are 1 and 3. Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<1\) or \(x>3\). Since given that \(x\) is positive then only one range is valid: \(x>3\). Sufficient. (2) (x  2)^2 > 9 > \((x+1)(x5)>0\) > roots are 1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<1\) or \(x>5\). Since given that \(x\) is positive then only one range is valid: \(x>5\). Sufficient. Answer: D. Solving inequalities: x24x94661.html#p731476 ( check this one first) inequalitiestrick91482.htmldatasuffinequalities109078.htmlrangeforvariablexinagiveninequality109468.html?hilit=extreme#p873535everythingislessthanzero108884.html?hilit=extreme#p868863Hope it helps. Another approach:If x is positive, is x > 3 ?(1) (x  1)^2 > 4 > since both sides of the inequality are nonnegative then we can take square root from both parts: \(x1>2\). \(x1\) is just the distance between 1 and \(x\) on the number line. We are told that this distance is more than 2: (1)13 so, \(x<1\) or \(x>3\). Since given that \(x\) is positive then only one range is valid: \(x>3\). Sufficient. (2) (x  2)^2 > 9 > \(x2>3\). The same here: \(x2\) is just the distance between 2 and \(x\) on the number line. We are told that this distance is more than 3: (1)25 so, \(x<1\) or \(x>5\). Since given that \(x\) is positive then only one range is valid: \(x>5\). Sufficient. Answer: D. I used x= +6 and 6 ..Which is true in both the cases..it shud be E.. Stem says that x is a positive number, thus x cannot be 6. Hope it's clear.
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Re: If x is positive, is x > 3
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27 Oct 2013, 22:43
Bunuel wrote: If x is positive, is x > 3 ?
(1) (x  1)^2 > 4 > \((x+1)(x3)>0\) > roots are 1 and 3. Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<1\) or \(x>3\). Since given that \(x\) is positive then only one range is valid: \(x>3\). Sufficient.
(2) (x  2)^2 > 9 > \((x+1)(x5)>0\) > roots are 1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<1\) or \(x>5\). Since given that \(x\) is positive then only one range is valid: \(x>5\). Sufficient.
Answer: D.
I have a question if the sign was "<" instead of ">" what would the solution be? (x+1)(x3) < 0
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If x is positive, is x > 3
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27 Oct 2013, 22:57
fozzzy wrote: Bunuel wrote: If x is positive, is x > 3 ?
(1) (x  1)^2 > 4 > \((x+1)(x3)>0\) > roots are 1 and 3. Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<1\) or \(x>3\). Since given that \(x\) is positive then only one range is valid: \(x>3\). Sufficient.
(2) (x  2)^2 > 9 > \((x+1)(x5)>0\) > roots are 1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<1\) or \(x>5\). Since given that \(x\) is positive then only one range is valid: \(x>5\). Sufficient.
Answer: D.
I have a question if the sign was "<" instead of ">" what would the solution be? (x+1)(x3) < 0 \((x+1)(x3)<0\) > \(1<x<3\). \((x+1)(x5)<0\) > \(1<x<5\).
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Re: If x is positive, is x > 3
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28 Dec 2015, 02:46
Key thing to note is that X is positive. Statement 1  If (x1)^2 > 4, x has to be greater than 3. If x = 1, 2 or 3, then the expression will only be < or equal to 4. Hence 1 is sufficient. Statement 2  If (x  2)^2 > 9, x has to be greater than 3. If x = 1, 2 or 3, then the expression will not be > 9. Hence 2 is also sufficient. Answer  D. Bunuel  Is this really a 700 level question ?



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Re: If x is positive, is x > 3
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06 Jul 2018, 00:41
rohitgoel15 wrote: If x is positive, is x > 3 ?
(1) (x  1)^2 > 4 (2) (x  2)^2 > 9
Given x > 0, asked is x > 3? Statement 1: (x  1)^2 > 4 Hence lx  1l > 2, so we have x > 3 or x < 1 Since x > 0, there fore x > 3. Answer is YES. Statement 1 alone is Sufficient. Statement 2: (x  2)^2 > 9 Hence lx  2l > 3, so we have x > 5 or x < 1 Since x > 0, therefore x > 5 & hence x > 3. Answer is YES. Statement 2 alone is Sufficient. Answer D. Thanks, GyM
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Re: If x is positive, is x > 3
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22 Jul 2018, 06:46
Bunuel niks18 pushpitkc chetan2u KarishmaB amanvermagmat pikolo2510I understood alternate approach by Bunuel since taking square root of a squared variable and turning it in to absolute value came more naturally to me. Where I faltered is doing this is less than 2 mins as per below steps: x1 > 2 x1 >2 or x+1 > 2 x>3 or x>1 ie x<1 Similar painstaking calculations I had to do for St 2
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If x is positive, is x > 3
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22 Jul 2018, 07:41
adkikaniremember a small rule and this question can be done in 30 seconds. See my approach. Attachment:
IMG20180722WA0005.jpg [ 84.46 KiB  Viewed 2847 times ]
adkikani wrote: Bunuel niks18 pushpitkc chetan2u KarishmaB amanvermagmat pikolo2510I understood alternate approach by Bunuel since taking square root of a squared variable and turning it in to absolute value came more naturally to me. Where I faltered is doing this is less than 2 mins as per below steps: x1 > 2 x1 >2 or x+1 > 2 x>3 or x>1 ie x<1 Similar painstaking calculations I had to do for St 2
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Re: If x is positive, is x > 3
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22 Jul 2018, 08:24
Your approach is perfect . Practice more of such questions so that the steps come like a reflex and save time. adkikani wrote: Bunuel niks18 pushpitkc chetan2u KarishmaB amanvermagmat pikolo2510I understood alternate approach by Bunuel since taking square root of a squared variable and turning it in to absolute value came more naturally to me. Where I faltered is doing this is less than 2 mins as per below steps: x1 > 2 x1 >2 or x+1 > 2 x>3 or x>1 ie x<1 Similar painstaking calculations I had to do for St 2
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