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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, 07:11

Statement 1: If x has to be an integer only x=2 would yield a number that is less than 0. Sufficient. Statement 2: x can take on any number of integers that would yield a number greater than 0. Not Sufficient.

Re: If x is an integer, what is the value of x? (1) x^2 - 4x + 3 [#permalink]

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07 Jul 2013, 21:20

1

This post received KUDOS

(1) reduces to (x-3)(x-1) < 0 which is possible only when x-3 and x-1 are of the opposite signs which => 1<x<3. Since x is an integer, we have a solution for x=2.

(2) reduces to (x+3)(x+1) > 0 which is possible for x < -3 or x > -1 which has many solutions, Not sufficient.

Thanks Bunuel, The second link with the graphical approach made everything very clear. I have been struggling to figure out how this works. Thanks again.

Re: If x is an integer, what is the value of x? (1) x^2 - 4x + 3 [#permalink]

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24 Aug 2014, 12:19

Hi Bunuel, Can I safely deduce the following

When (x+3)(x+1) = 0 , There are exactly two solutions

When (x+3) (x+1)> 0 , there are multiple solutions. Because by putting any positive value of x I can get >0

When (x+3) (x+1)<0, Either (x+3) or (x+1) has to be negative and obviously none can be Zero. In that case to make negative I can take only one value which is -2.

I don't want to concentrate on graphical method. Will the approach work for these types of questions always? Please help

When (x+3)(x+1) = 0 , There are exactly two solutions

When (x+3) (x+1)> 0 , there are multiple solutions. Because by putting any positive value of x I can get >0

When (x+3) (x+1)<0, Either (x+3) or (x+1) has to be negative and obviously none can be Zero. In that case to make negative I can take only one value which is -2.

I don't want to concentrate on graphical method. Will the approach work for these types of questions always? Please help

That's not correct.

(x + 3)(x + 1) > 0 means that x < -3 or x > -1. (x + 3)(x + 1) < 0 means that -3 < x < -1.
_________________

Re: If x is an integer, what is the value of x? (1) x^2 - 4x + 3 [#permalink]

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24 Aug 2014, 12:42

Bunuel wrote:

Raihanuddin wrote:

Hi Bunuel, Can I safely deduce the following

When (x+3)(x+1) = 0 , There are exactly two solutions

When (x+3) (x+1)> 0 , there are multiple solutions. Because by putting any positive value of x I can get >0

When (x+3) (x+1)<0, Either (x+3) or (x+1) has to be negative and obviously none can be Zero. In that case to make negative I can take only one value which is -2.

I don't want to concentrate on graphical method. Will the approach work for these types of questions always? Please help

That's not correct.

(x + 3)(x + 1) > 0 means that x < -3 or x > -1. (x + 3)(x + 1) < 0 means that -3 < x < -1.

So for the 1st one value can be -4, -5....... or can be 0, 1, 2,...... and can be fraction also. For the 2nd one only -2 if integer but can any number in between -1 and and -3 if not integer. Actually in the original question I have seen integer. That's why I picked integer. If x is integer, is my reasoning correct to find answer? So far I think I am saying the same thing you have given for the above two.

Re: If x is an integer, what is the value of x? (1) x^2 - 4x + 3 [#permalink]

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09 Oct 2014, 04:54

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) \(x^2-4x+3<0\) --> \(1<x<3\) --> 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.

Hello Bunuel, I am not sure if I followed your explanation all that clearly. I was able to get to the factors of the equation but after that, you mention about "<" symbol and as a result you arrive at \(1<x<3\). How come? I tried to grasp your graphical approach but that did sit well either. Could you please elaborate? Thank you in advance!

(1) \(x^2-4x+3<0\) --> \(1<x<3\) --> 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.

Hello Bunuel, I am not sure if I followed your explanation all that clearly. I was able to get to the factors of the equation but after that, you mention about "<" symbol and as a result you arrive at \(1<x<3\). How come? I tried to grasp your graphical approach but that did sit well either. Could you please elaborate? Thank you in advance!

If x is an integer, what is the value of x? (1) x^2 - 4x + 3 [#permalink]

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09 Oct 2014, 11:36

Blackbox 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) \(x^2-4x+3<0\) --> \(1<x<3\) --> 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.

Hello Bunuel, I am not sure if I followed your explanation all that clearly. I was able to get to the factors of the equation but after that, you mention about "<" symbol and as a result you arrive at \(1<x<3\). How come? I tried to grasp your graphical approach but that did sit well either. Could you please elaborate? Thank you in advance!

After spending long time on this topic, I have come to an easy solution for me. You may like it

(1) (x-3)(x-1)<0. Now use the logic. To get negative result Either (x-3) or (x-1) has to be negative and the other has to +ve

Here, (x-3) can't be positive, Because if x-3>0 then x>3 and if x>3 then x-1 will become +ve. So (x-3)(x-1) will become +ve.

Therefore, (x-3) has to be negative,x-3<0 or x<3 and and as I said earlier x-1>0 or x>1

Finally we get, 1<x<3. Sufficient because 2 is the only integer between 1 and 3.

(2) we don't need to find equation. Just by putting multiple positive values we can satisfy statement 2. So Not sufficient

Ans. A Hopefully my thinking is correct. I expect expert reply. Can I get more such question to practice?

Thanks bub. I am good with the method to solve inequalities. Do you by any chance have more questions related to Q. Inequalities, please? I've come across four such questions so far on this forum, but again, I have not done a search for them yet ... which I will do now.

Thanks bub. I am good with the method to solve inequalities. Do you by any chance have more questions related to Q. Inequalities, please? I've come across four such questions so far on this forum, but again, I have not done a search for them yet ... which I will do now.

Re: If x is an integer, what is the value of x? (1) x^2 - 4x + 3 [#permalink]

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10 Oct 2014, 04:03

Bunuel 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) \(x^2-4x+3<0\) --> \(1<x<3\)--> 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.[/quote]

With reference to your statement \(1<x<3\) above, I calculated the range as \(x<1\) or \(x<3\) What I did wrong?[/quote]

How to solve quadratic inequalities - Graphic approach.

\(x^2-4x+3<0\) is the graph of parabola and it look likes this:

Attachment:

en.plot (1).png

Intersection points are the roots of the equation \(x^2-4x+3=0\), which are \(x_1=1\) and \(x_2=3\). "<" sign means in which range of \(x\) the graph is below x-axis. Answer is \(1<x<3\) (between the roots).

If the sign were ">": \(x^2-4x+3>0\). First find the roots (\(x_1=1\) and \(x_2=3\)). ">" sign means in which range of \(x\) the graph is above x-axis. 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: \(-x^2-x+12>0\), first rewrite this as \(x^2+x-12<0\) (so that the coefficient of x^2 to be positive. It's possible to solve without rewriting, but easier to master one specific pattern).

\(x^2+x-12<0\). Roots are \(x_1=-4\) and \(x_1=3\) --> below ("<") the x-axis is the range for \(-4<x<3\) (between the roots).

Again if it were \(x^2+x-12>0\), 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).

Re: If x is an integer, what is the value of x? (1) x^2 - 4x + 3 [#permalink]

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02 Jul 2015, 11:34

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) \(x^2-4x+3<0\) --> \(1<x<3\) --> 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.

I understand this but I have problem when I start plugging values in 1. when x =0 => 0 -0+ 3 < 0 implies NO when x=1 => 1-4 +3 < 0 NO when x =2 => 4 -8 +3 =-1 < 0 is YES since we can have multiple values of x, isn't that statement should be insufficient.

gmatclubot

Re: If x is an integer, what is the value of x? (1) x^2 - 4x + 3
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02 Jul 2015, 11:34

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