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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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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,
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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 (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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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.

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

Solving and Factoring Quadratics:
https://www.purplemath.com/modules/solvquad.htm
https://www.purplemath.com/modules/factquad.htm

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]
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]
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: (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

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

Hi Bunuel,
Great post.
Could you please let explain why do we not consider the possibility of x=1 or x=3 in the expression x^2-4x+3<0, well now I know when we plug in values x=1/3 the expression will become zero and won't hold true.
However I have seen few examples wherein =</=> possibilities are considered.

Is there any technique for this?

KarishmaB Bunuel

Thank you,
Prachita
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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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Prachita123 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.

Hi Bunuel,
Great post.
Could you please let explain why do we not consider the possibility of x=1 or x=3 in the expression x^2-4x+3<0, well now I know when we plug in values x=1/3 the expression will become zero and won't hold true.
However I have seen few examples wherein =</=> possibilities are considered.

Is there any technique for this?

KarishmaB Bunuel

Thank you,
Prachita

Prachita123 - It depends on what is given to you.

We are given in stmnt 1

$$(1) x^2 - 4x + 3 < 0$$
So we know that x lies between 1 and 3. It can take a value 2 for example.
$$2^2 - 4*2 + 3 < 0$$
$$-1 < 0$$
(True)

But can x be 1?
$$1^2 - 4*1 + 3 < 0$$
$$0 < 0$$
(Not true)

Hence x cannot be 1.

On the other hand, if the given inequality to us was
$$(1) x^2 - 4x + 3 \leq 0$$

Now this would hold for x = 1 as well as 3 and all numbers between 1 and 3 because both LHS equal to 0 and LHS less than 0 are allowed.
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Re: If x is an integer, what is the value of x? (1) x^2 - 4x + 3 [#permalink]
I am guessing, logically thinking, we don't even need to solve for St 2, as we know it would lead to a split inequality due to the "greater than" sign and there are not enough "restrictive conditions" in the stem, for it to produce a single integer value?
I wonder if there are any equations possible at all, under which St 2 could have produced a single integer value, for a question like this?

Bunuel
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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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TargetMBA007 wrote:
I am guessing, logically thinking, we don't even need to solve for St 2, as we know it would lead to a split inequality due to the "greater than" sign and there are not enough "restrictive conditions" in the stem, for it to produce a single integer value?
I wonder if there are any equations possible at all, under which St 2 could have produced a single integer value, for a question like this?

Bunuel

The upward-facing parabola will always yield infinitely many values of x that satisfy the '> 0' condition. Similarly, the downward-facing parabola will always yield infinitely many values of x that satisfy the '< 0' condition.
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