How many integrals value of x satisfy the inequality (1-x2)(4-x2)(9-x2

Question

How many integrals value of x satisfy the inequality (1-x^2)(4-x^2)(9-x^2) > 0 ?

A. 0
B. 1
C. 3
D. 5
E. Greater than 5

CesarAMSTech · Answer

I will fully factorise the expresion as shown in the picture attached and then i will draw each of the critic points (the points that make the whole expression equal to zero) over the number line and analyze the sign of the regions between each couple.

Based on that, its clear that the unique integer value included in the positive region (so that the expression >= 0 ) is right in the middle of the number line, i mean 0.

Answer:   B) One value   (Zero)

firas92 · Answer

(1-x^2)(4-x^2)(9-x^2) = (1+x)(1-x)(2+x)(2-x)(3+x)(3-x)

If x<-3 or x>3,

(1+x)(1-x)(2+x)(2-x)(3+x)(3-x)<0

If x=-3,-2,-1,1,2 or 3

(1+x)(1-x)(2+x)(2-x)(3+x)(3-x)=0

If x=0,

(1+x)(1-x)(2+x)(2-x)(3+x)(3-x)=36

So only 1 integer value of x is possible

Answer is B

Posted from my mobile device

TestPrepUnlimited · Answer

(1-x^2)(4-x^2)(9-x^2) has 6 zero solutions, two for each bracket we have currently. They are -3, -2, -1, 1, 2, 3. Let's start from when x > 3, we would have (-) * (-) * (-) = (-), a negative result so none of the integers bigger than 3 satisfy this inequality. The same applies for x < -3. Another quick way to see this is, each zero solution flips the sign of the result. We can start from the x > 3 case which gives a negative result, and to reach x < -3 we need to pass 6 zero solutions. Then the sign of (1-x^2)(4-x^2)(9-x^2) would still remain negative after 6 flips. Then we only have x = -3 through x = 3 as possible solutions. None of the zero's will give a positive result so we can only try x = 0. Plugging in that gives 1 * 4 * 9 > 0, our only solution. Ans: B

GMATinsight · Answer

(1-x^2)(4-x^2)(9-x^2) > 0

The critical values are x (for which function is zero) = {-3, -2, -1, 1, 2, 3}

All these values of  consecutive integers  except integer 0 among them so effectively we have 3 ranges

Range 1: x < -3
Range 2: x = 0
Range 3: x > 3

Only x = 0 satisfies for the function to be positive hence

Answer: Option B

Kritisood · Answer

to use the wavy line approach we need to get the factors in the form of  (x+a) or (x-a) . this can be done by taking -1 common.
(x^2-1)(x^2-4)(x^2-9) < 0 (flip the sign as multiplying by -1 on both sides)

draw the wavy line at points -3,-2,-1,1,2,3
only 0 is the integral value.

robinbjoern · Answer

I don't get why you start with negative on the RHS on your number line. https://www.youtube.com/watch?v=j-jGsD2dweI In this video it is clearly shown that we have to start with + on the right hand side, when solving those inequality.

ShreyasJavahar · Answer

That approach doesn't always work. A more efficient way would be to first identify all the zeroes/critical points of the equation, then pick the largest range available from these critical points (in this case that range is x>3), substitute a relevant value, say 4, into the equation and see whether the resulting value is greater than 0 or lesser. If it is lesser than 0, start with a negative from the RHS and alternate the signs, based on the powers of roots, as you traverse left; if it is greater than 0, do the opposite, start with a positive from the RHS.

mysterymanrog · Answer

We can check zero value:
x=0 at the points:
x=-3,-2,-1,1,2,3

For x>3 -> (-)(-)(-)<0
For 2<x<3 -> (-)(-)(+)>0 -> no integers between 2 and 3
for 1<x<2 (-)(+)(+)<0
for -1<x<1 (+)(+)(+)>0 -> only integer is x=0
for -2<x<-1 (-)(+)(+)<0
for -3<x<-2 (-)(-)(+)>0 -> no integers between -3 and -2
for x<-3 (-)(-)(-)<0

only 1 solution, x=0
B

Can,Will · Answer

egmat   KarishmaB   gmatophobia  Kindly explain whats wrong with my approach

gmatophobia · Answer

You have performed each computation in isolation and have ignored the fact that the obtained values for one computation can make the other term zero. For example, for the second term,  (4 - x^2) > 0 , we have obtained that the valid values can be -1, 0, and 1. However, among these three values, only 0 is valid for the inequality to hold true. This is because both -1 and 1 result in  (1-x^2) = 0   and the inequality then doesn't hold true. For the approach you've used, you should take the common terms.

P.S. There is also a broader issue with this approach. It may not be applicable to this question, but thought of sharing it in general. The only criteria that you've used is when all three terms are positive. However, we can have other possible cases as well. Ex: when two terms are negative and one is positive.

shubhim20 · Answer

how did we get the range?

Bunuel · Answer

Check the highlighted links below for the trick described above:

9. Inequalities 
 
    Theory    
 
 Inequalities Made Easy! 
 Inequalities: Tips and hints 
 Arithmetic with Inequalities 
  Wavy Line Method Application - Complex Algebraic Inequalities 
  Solving Quadratic Inequalities: Graphic Approach 
  Inequalities - Quadratic Inequalities 
 Graphic approach to problems with inequalities 
 Inequalities trick 
  INEQUATIONS(Inequalities) Part 1 
 INEQUATIONS(Inequalities) Part 2 
 
    Questions    
 
 Inequality and absolute value questions from my collection 
 DS Questions 
 PS Questions

For more check  Ultimate GMAT Quantitative Megathread

Hope it helps.

rajkoushik17 · Answer

Without expanding the equation much it is clear that the LHS will be positive if two of the terms are -ve or all the three are +ve. If two are to be -ve then 4<x2<9. Since we need integral value of X there is no solution for X in this approach. If all the three terms are to be positive, X2<1 only 0 is the possible solution.

Kritisood

How many integrals value of x satisfy the inequality (1-x^2)(4-x^2)(9-x^2) > 0 ?

A. 0
B. 1
C. 3
D. 5
E. Greater than 5

Kinshook · Answer

How many integrals value of x satisfy the inequality (1-x^2)(4-x^2)(9-x^2) > 0 ?

(1-x^2)(4-x^2)(9-x^2) > 0
(x^2-1)(x^2-4)(x^2-9) < 0
(x-1)(x+1)(x-2)(x+2)(x-3)(x+3) < 0

Using wavy curve method,

-3,-2,-1,1,2,3

2 < x < 3
-1 < x < 1
-3 < x < -2

x = 0 is the only solution

IMO B

How many integrals value of x satisfy the inequality (1-x2)(4-x2)(9-x2

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How many integrals value of x satisfy the inequality (1-x2)(4-x2)(9-x2

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