How to solve quadratic inequalities - Graphic approach.

\(x^2-4x+3<0\) is the graph of parabola and it look likes this:
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,

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.
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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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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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Great explanation Bunuel!! Kudos!
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Thanks for the explanation.

Kudos Bunuel. I struggled with these questions always. But now its clear.

You are a genius!!! I have been struggling with inequalities particularly quadratics, this solved my issue!!

Amazing explanation.... wow.

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

Hope it helps.
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Nice little trick! And useful too! This combined with Bunuel's graphical approach to quadratic inequalities make these problem types so easy now..

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

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

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

(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.

Thus (A) is the answer.

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


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