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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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If x is an integer, what is the value of x?

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

Originally posted by dimitri92 on 23 May 2010, 00:31.
Last edited by Bunuel on 04 Aug 2012, 00:59, edited 1 time in total.
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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) $$x^2-4x+3<0$$ --> $$1<x<3$$--> 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.

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.

$$x^2-4x+3<0$$ is the graph of parabola and it look likes this:
Attachment: en.plot (1).png [ 3.79 KiB | Viewed 90264 times ]

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.
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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: (x-1)(x-2)(x-3)(x-7) < 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,
##### General Discussion
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Joined: 02 Sep 2009
Posts: 59561
Re: x2 - 4x + 3 < 0  [#permalink]

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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.

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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 (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.
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GMAT 1: 680 Q48 V34 Re: x2 - 4x + 3 < 0  [#permalink]

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

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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GMAT 1: 680 Q48 V34 Re: x2 - 4x + 3 < 0  [#permalink]

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Great explanation Bunuel!! Kudos!
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Re: x2 - 4x + 3 < 0  [#permalink]

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Thanks for the explanation.
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Re: x2 - 4x + 3 < 0  [#permalink]

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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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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.

$$x^2-4x+3<0$$ 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.
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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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Amazing explanation.... wow.
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Posts: 59561
Re: x2 - 4x + 3 < 0  [#permalink]

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kartik222 wrote:
@Bunuel,
I don't understand how you calculated the roots this equation: $$x^2+x-12<0$$. your answer: Roots are $$x_1=-4$$ and $$x_1=3$$

$$x^2+x-12<0$$
(x-4)(x+3)<0
so, two point now will be 4 and -3

Now the we can plot the graph and get to the answer.

Thanks!

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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1
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: (x-1)(x-2)(x-3)(x-7) < 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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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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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: (x-1)(x-2)(x-3)(x-7) < 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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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:

(x-1)(x-2)(x-3)(x-7) < 0
= (10-1)(10-2)(10-3)(10-7)
You get 9*8*7*3 which means the expression is positive when x =10
The roots of the expression (x-1)(x-2)(x-3)(x-7) < 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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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) (x-1)(x-3) < 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

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(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.

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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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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?
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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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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

_________________ Re: If x is an integer, what is the value of x? (1) x^2 - 4x + 3   [#permalink] 17 May 2014, 05:45

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