A. (x+3)(x+2)>0 using FOIL
B. (x+2)(x+2)>16 by squaring both sides
C. x<7/2 by completing inequality
D. (x-3)(x-1)<0 using FOIL
E.

in A we have x>-3 and x>-2 OR x<-3 and x<-2, solution area/range x>-2
in B, x>some +ve value
in C, just infinite range
in D, x<3 and x>1 OR x>3 and x<1, solution area 1<x<3

d

bharti.2010 wrote:Which of the following inequalities have a finite range of values of "x" satisfying them?
A. x2 + 5x + 6 > 0
B. |x + 2| > 4
C. 9x - 7 < 3x + 14
D. x2 - 4x + 3 < 0
E. (B) and (D)

I am really weak in Inequality. Please help me with this question and suggest me some good study material(with practice questions) for this topic and Permutation, combination & Probibility.

Hi Bharti,

solving inequalities is about one thing: sign changes. Find all the points at
which there are sign changes - we call these points critical values. Then determine
which, if any, of the intervals bounded by these critical values result in a solution. The
solution to the inequality will consist of the set of all points contained by the solution
intervals.

Which of the following inequalities have a finite range of values of "x" satisfying them?
A. x2 + 5x + 6 > 0
B. |x + 2| > 4
C. 9x - 7 < 3x + 14
D. x2 - 4x + 3 < 0
E. (B) and (D)

A. x2 + 5x + 6 > 0 =>(x+2)(x+3)>0

Here the critical points are when the expression (x+2)(x+3) = 0, i.e. at x = -2 and -3.
Let us check if the intervals bounded by these critical values result in a solution.

If x<-3
Expression (x+2)*(x+3)>0 (if x =-4, (-4+2)*(-4+3) >0) - you can stop solving the problem here because x<-3 is the solution and consists of an infinite solutions.
If -3<x<-2,
Expression (x+2)*(x+3)<0 (if x =-2.5, (-2.5+2)*(-2.5+3)<0)
If x<-2
Expression (x+2)*(x+3)>0 (if x =0, (0+2)*(0+3) >0).

Solution for the inequality is x<-3(-Infinity to -3 in school terms) or x>-2(2 to infinity in school terms), which is NOT a finite set of real numbers.

B
|x + 2| > 4.
If|Y|>a then the solution is Y > a or Y <-a
So,
x + 2 > 4 or x + 2 < -4
x > 2 (2 to Infinity in school terms)or x < -6 (-Infinity to -6 in school terms)is the solution, which is NOT a finite set of real numbers.

C

9x - 7 < 3x + 14
6x < 21
x < 7/2 (-Infinity to 7/2 as learnt at school)is the solution, which is NOT a finite set of real numbers.

D

You need not check for this option because this is the only one left but let me check for your benefit.

x2 - 4x + 3 < 0 , (x-1)(x-3)<0
Here the critical points are when the expression (x-1)(x-3) = 0, i.e. at x = 1 and 3.
Let us check if the intervals bounded by these critical values result in a solution.

If x<1
Expression (x-1)(x-3)>0 (if x=0, (0-1)*(0-3) >0)
If 1<x<3,
Expression (x-1)(x-3)<0 (if x=2, (2-1)*(2-3)<0)
If x>3
Expression (x-1)(x-3)>0 (if x=4, (4-1)*(4-3) >0)

1<x<3 is the solution, which is a finite set.

Option D, is the answer!

Thanks Neel for your good explanation. But I still don't get explantion for A and B.

This concept is not clear to me:
A. x2 + 5x + 6 > 0 =>(x+2)(x+3)>0

In this case, (x+2) and (x+3) both are either +ve or -ve. [+ve * +ve >0, -ve * -ve> 0]

If both are +ve then x> -2, -3
and if -ve then x< -2, -3.

What to do next is not clear to me?

i believe Nil's was far away stretching any detail in his post

let me get hold of this query - you must supply (plug in) your own values within the given ranges

If both are +ve then x> -2, -3
and if -ve then x< -2, -3.

and decide which one is the solid range for the inequality, that is you select any number within given ranges and see if it satisfies both expressions in the parentheses (here, for this case). If there's no violation of sign after plug-in then you keep that range. As you see options A and B show infinite ranges towards + infinity.

bharti.2010 wrote:Thanks Neel for your good explanation. But I still don't get explantion for A and B.

This concept is not clear to me:
A. x2 + 5x + 6 > 0 =>(x+2)(x+3)>0

In this case, (x+2) and (x+3) both are either +ve or -ve. [+ve * +ve >0, -ve * -ve> 0]

If both are +ve then x> -2, -3
and if -ve then x< -2, -3.

What to do next is not clear to me?