Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
(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 70579 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]
Show Tags
24 May 2012, 07:54
12
6
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
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]
Show Tags
09 Jun 2012, 03:24
1
1
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..!!!
Re: If x is an integer, what is the value of x? (1) x^2 - 4x + 3 [#permalink]
Show Tags
03 Aug 2012, 23:52
3
3
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..
close
50% (01:11) correct 
