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If x is an integer, what is the value of x? (1) x^2  4x + 3 [#permalink]
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23 May 2010, 00:31
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If x is an integer, what is the value of x? (1) x^2  4x + 3 < 0 (2) x^2 + 4x +3 > 0
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Last edited by Bunuel on 04 Aug 2012, 00:59, edited 1 time in total.
OA added.



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Re: x2  4x + 3 < 0 [#permalink]
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23 May 2010, 01:07
C
St. 1: insuff > x< 3 or x < 1
St. 2: insuff > x> 3 or x > 1
St. 1 & 2 : Sufficient > 1 < x < 1 and as x is an integer, x = 0.
please post the OA.



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Re: x2  4x + 3 < 0 [#permalink]
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Re: What is the value of x? [#permalink]
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Hussain15 wrote: If \(x\) is an integer, what is the value of \(x\) ?
(1) \(x^2  4x + 3 < 0\) (2) \(x^2 + 4x +3 > 0\) IMO A 1st equation gives (x1)*(x3) <0 which is only possible when x=2 2nd gives (x+1)*(x+3) > 0 which is possible when x< 3 and x> 1 thus multiple values. Thus A OA pls.
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Re: x2  4x + 3 < 0 [#permalink]
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30 May 2010, 04:27
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Bunuel wrote: dimitri92 wrote: If x is an integer, what is the value of x?
(1) x2  4x + 3 < 0 (2) x2 + 4x +3 > 0 Given: \(x=integer\). (1) \(x^24x+3<0\) > \( 1<x<3\)> as \(x\) is an integer then \(x=2\). Sufficient. (2) \(x^2+4x+3>0\) > \(x<3\) or \(x>1\) > multiple values are possible for integer \(x\). Not sufficient. Answer: A. With reference to your statement \(1<x<3\) above, I calculated the range as \(x<1\) or \(x<3\) What I did wrong?
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Re: x2  4x + 3 < 0 [#permalink]
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Hussain15 wrote: Bunuel wrote: dimitri92 wrote: If x is an integer, what is the value of x?
(1) x2  4x + 3 < 0 (2) x2 + 4x +3 > 0 Given: \(x=integer\). (1) \(x^24x+3<0\) > \( 1<x<3\)> as \(x\) is an integer then \(x=2\). Sufficient. (2) \(x^2+4x+3>0\) > \(x<3\) or \(x>1\) > multiple values are possible for integer \(x\). Not sufficient. Answer: A. With reference to your statement \(1<x<3\) above, I calculated the range as \(x<1\) or \(x<3\) What I did wrong? How to solve quadratic inequalities  Graphic approach.\(x^24x+3<0\) is the graph of parabola and it look likes this: Attachment:
en.plot (1).png [ 3.79 KiB  Viewed 65535 times ]
Intersection points are the roots of the equation \(x^24x+3=0\), which are \(x_1=1\) and \(x_2=3\). "<" sign means in which range of \(x\) the graph is below xaxis. Answer is \(1<x<3\) (between the roots). If the sign were ">": \(x^24x+3>0\). First find the roots (\(x_1=1\) and \(x_2=3\)). ">" sign means in which range of \(x\) the graph is above xaxis. Answer is \(x<1\) and \(x>3\) (to the left of the smaller root and to the right of the bigger root). This approach works for any quadratic inequality. For example: \(x^2x+12>0\), first rewrite this as \(x^2+x12<0\) (so that the coefficient of x^2 to be positive. It's possible to solve without rewriting, but easier to master one specific pattern). \(x^2+x12<0\). Roots are \(x_1=4\) and \(x_1=3\) > below ( "<") the xaxis is the range for \(4<x<3\) (between the roots). Again if it were \(x^2+x12>0\), then the answer would be \(x<4\) and \(x>3\) (to the left of the smaller root and to the right of the bigger root). Hope it helps.
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Re: x2  4x + 3 < 0 [#permalink]
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30 May 2010, 23:01



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Re: x2  4x + 3 < 0 [#permalink]
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31 May 2010, 04:38
Thanks for the explanation.



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Re: x2  4x + 3 < 0 [#permalink]
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15 Mar 2011, 20:04
From (1) we have (x3)(x1) < 0 => 3 < x < 1, and there is only one integer between 1 and 3 which is 2, so (1) is sufficient. From (2) we have (x+1)(x+3) > 0 So either x > 1 or x <  3, which is not enough Hence the answer is A
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Re: x2  4x + 3 < 0 [#permalink]
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25 Jun 2011, 00:04
Kudos Bunuel. I struggled with these questions always. But now its clear.
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Re: x2  4x + 3 < 0 [#permalink]
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20 Apr 2012, 10:15
You are a genius!!! I have been struggling with inequalities particularly quadratics, this solved my issue!! Bunuel wrote: Hussain15 wrote: Bunuel wrote: How to solve quadratic inequalities  Graphic approach.\(x^24x+3<0\) is the graph of parabola and it look likes this: Attachment: en.plot (1).png Intersection points are the roots of the equation \(x^24x+3=0\), which are \(x_1=1\) and \(x_2=3\). "<" sign means in which range of \(x\) the graph is below xaxis. Answer is \(1<x<3\) (between the roots). If the sign were ">": \(x^24x+3>0\). First find the roots (\(x_1=1\) and \(x_2=3\)). ">" sign means in which range of \(x\) the graph is above xaxis. Answer is \(x<1\) and \(x>3\) (to the left of the smaller root and to the right of the bigger root). This approach works for any quadratic inequality. For example: \(x^2x+12>0\), first rewrite this as \(x^2+x12<0\) (so that the coefficient of x^2 to be positive. It's possible to solve without rewriting, but easier to master one specific pattern). \(x^2+x12<0\). Roots are \(x_1=4\) and \(x_1=3\) > below ( "<") the xaxis is the range for \(4<x<3\) (between the roots). Again if it were \(x^2+x12>0\), then the answer would be \(x<4\) and \(x>3\) (to the left of the smaller root and to the right of the bigger root). Hope it helps.



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Re: If x is an integer, what is the value of x? (1) x^2  4x + 3 [#permalink]
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23 Apr 2012, 14:10
Amazing explanation.... wow.



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Re: x2  4x + 3 < 0 [#permalink]
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12 May 2012, 11:09



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Re: If x is an integer, what is the value of x? (1) x^2  4x + 3 [#permalink]
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24 May 2012, 07:54
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Hi,
I won't mind drawing the parabola for a quadratic equation. But a better approach is to check the sign of expression on a number line.
for example: (x1)(x2)(x3)(x7) < 0
To check the intervals in which this inequality holds true, we need to pick only one value from the number line. Lets say x = 10, then (9)(8)(7)(3) > 0, in every alternate interval the sign would be + for the above expression
(+)1()2(+)3()7(+)
Thus, inequality would hold true in the intervals: 1 < x < 2 3 < x < 7
This is the general approach which can be used when you see multiplications in inequalities.
Regards,



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Re: If x is an integer, what is the value of x? (1) x^2  4x + 3 [#permalink]
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09 Jun 2012, 03:24
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cyberjadugar wrote: Hi,
I won't mind drawing the parabola for a quadratic equation. But a better approach is to check the sign of expression on a number line.
for example: (x1)(x2)(x3)(x7) < 0
To check the intervals in which this inequality holds true, we need to pick only one value from the number line. Lets say x = 10, then (9)(8)(7)(3) > 0, in every alternate interval the sign would be + for the above expression
(+)1()2(+)3()7(+)
Thus, inequality would hold true in the intervals: 1 < x < 2 3 < x < 7
This is the general approach which can be used when you see multiplications in inequalities.
Regards, Nice little trick! And useful too! This combined with Bunuel's graphical approach to quadratic inequalities make these problem types so easy now..



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Re: If x is an integer, what is the value of x? (1) x^2  4x + 3 [#permalink]
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03 Aug 2012, 18:59
Wow. I've been struggling for hours on this.
This is the best explanation I've seen of any GMAT related stuff ever.
So cool!!!!!!!!!! Thanks



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Re: If x is an integer, what is the value of x? (1) x^2  4x + 3 [#permalink]
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03 Aug 2012, 23:19
cyberjadugar wrote: Hi,
I won't mind drawing the parabola for a quadratic equation. But a better approach is to check the sign of expression on a number line.
for example: (x1)(x2)(x3)(x7) < 0
To check the intervals in which this inequality holds true, we need to pick only one value from the number line. Lets say x = 10, then (9)(8)(7)(3) > 0, in every alternate interval the sign would be + for the above expression
(+)1()2(+)3()7(+) Thus, inequality would hold true in the intervals: 1 < x < 2 3 < x < 7
This is the general approach which can be used when you see multiplications in inequalities.
Regards, Hi, Can u please eloaborate above highlighted part???. I m still confused how did u rearranged 9)(8)(7)(3) > 0 in to number line and subsequent results?? Help appriciated..!!!



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Re: If x is an integer, what is the value of x? (1) x^2  4x + 3 [#permalink]
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03 Aug 2012, 23:52
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bhavinshah5685 wrote: Hi, Can u please eloaborate above highlighted part???. I m still confused how did u rearranged 9)(8)(7)(3) > 0 in to number line and subsequent results?? Help appriciated..!!! Plug x = 10 for the expression: (x1)(x2)(x3)(x7) < 0 = (101)(102)(103)(107) You get 9*8*7*3 which means the expression is positive when x =10 The roots of the expression (x1)(x2)(x3)(x7) < 0 are: 1,2,3 and 7 Since 10 >7, the expression is positive when the value of x is greater than 7. From then on, just flip the sign every time you hit a root. So: from 3 to 7 the expression is ive from 2 to 3 the expression is +ive etc.. The idea is to start from the root with the highest absolute value, find out what the sign for the expression is and to work from there.. For more on this check out cyberjadugar's signature: Solving Inequalities



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Re: If x is an integer, what is the value of x? (1) x^2  4x + 3 [#permalink]
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07 Dec 2012, 04:51
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dimitri92 wrote: If x is an integer, what is the value of x?
(1) x^2  4x + 3 < 0 (2) x^2 + 4x +3 > 0 Using the amazing technique: (1) (x1)(x3) < 0 +(1)(3)+ (Select  curve if less than 0) Thus, 1 < x < 3 and only 2 is the integer in between 1 and 3. SUFFICIENT (2) (x+3)(x+1) > 0 +(3)(1)+ (Select + if greater than 0) x < 3 or x > 1 Thus, too many values of x INSUFFICIENT Answer: A
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Re: If x is an integer, what is the value of x? (1) x^2  4x + 3 [#permalink]
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17 Jan 2013, 05:44
dimitri92 wrote: If x is an integer, what is the value of x?
(1) x^2  4x + 3 < 0 (2) x^2 + 4x +3 > 0 1. (x3)(x1) < 0 Use the quick technique to getting roots: <(+)1()3(+)> 1 < x < 3, since x is an integer then x = 2 SUFFICIENT 2. (x+1)(x+3) > 0 <(+)(1)()(3)(+)> x < 1 or x > 3, too many possible roots, INSUFFICIENT. Answer: A
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