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(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?

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 43892 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).

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: (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

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:

(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..

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, 22:20

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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.

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).

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: (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

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

Show Tags

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: (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..!!!

gmatclubot

Re: If x is an integer, what is the value of x? (1) x^2 - 4x + 3
[#permalink]
03 Aug 2012, 23:19

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