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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 15439 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).

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]
24 May 2012, 06:54

2

This post received KUDOS

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]
09 Jun 2012, 02:24

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]
03 Aug 2012, 22:19

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..!!!

Re: If x is an integer, what is the value of x? (1) x^2 - 4x + 3 [#permalink]
03 Aug 2012, 22:52

2

This post received KUDOS

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..