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

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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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New post 17 May 2014, 05:45
gaurav1418z wrote:
Bunuel, very useful approach, thanks again

in the first equation,we wrote

(x-1)(x-3)<0 , so we know both terms have opposite signs, i.e

x>1 or x<3, since x is an integer, x has to be 2

However the other set is (x-1)<0 & (x-3)>0, or , x<1 or x>3, which can have many values, hence i marked statement 1 as insufficient,

for ex- x= -2 and y =4, equations becomes -3 * 1 = -3 , and it is still less than zero, hence satisfied.

Where am i going wrong?


How can x be simultaneously less than 1 and more than 3 for (x-1)(x-3)<0 ?

One of the ways of solving quadratic inequalities is given here: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476


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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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New post 16 Jun 2014, 04:15
hi.
Just wondering whether 1/0 will be an integer?
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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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New post 16 Jun 2014, 05:36
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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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New post 22 Jun 2014, 06:17
gurpreetsingh wrote:
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 (x-1)*(x-3) <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.


Hi Gurpreet ,

Can you please explain how did you did you say from the second statement that x <- 3and x> -1 ?

The factors of the equation would be -1 and -3 how how do we know the sign for x ... (I have read many forms but can't understand)
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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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New post 22 Jun 2014, 07:24
gauravsoni wrote:
gurpreetsingh wrote:
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 (x-1)*(x-3) <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.


Hi Gurpreet ,

Can you please explain how did you did you say from the second statement that x <- 3and x> -1 ?

The factors of the equation would be -1 and -3 how how do we know the sign for x ... (I have read many forms but can't understand)


Have you checked this post: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476 and this: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661-20.html#p1365336 ?
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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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New post 24 Aug 2014, 13:19
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
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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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New post 24 Aug 2014, 13:24
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.
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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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New post 09 Oct 2014, 05:54
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!
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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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New post 09 Oct 2014, 05:58
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!


Solving Quadratic Inequalities - Graphic Approach: solving-quadratic-inequalities-graphic-approach-170528.html
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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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New post 09 Oct 2014, 12:36
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?
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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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New post 10 Oct 2014, 05:03
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).

Hope it helps.[/quote]

I am fond of your explanations !!
++++kudos
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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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New post 02 Aug 2019, 22:34
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.


How are you getting X is greater than 1 in the first eqn. Shouldn't it be X is less than 1? The two factors are -3 & -1.
Am I going wrong somewhere?
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Re: If x is an integer, what is the value of x? (1) x^2 - 4x + 3   [#permalink] 02 Aug 2019, 22:34

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