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

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M7MBA: Moderator; Posts: 2058; Joined: Sun Oct 29, 2017 4:24 amFollowed by: 5 members; Thanked: 1 times

by M7MBA » Wed Jun 24, 2020 5:53 am

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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

[spoiler]OA=B[/spoiler]

Source: e-GMAT

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Source: Beat The GMAT — Problem Solving | Wed Jun 24, 2020 5:53 am

deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

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

by deloitte247 » Sun Jun 28, 2020 3:16 am

$$\left(1-x^2\right)\left(4-x^2\right)\left(9-x^2\right)>0$$
$$\left(1-x^2\right)=\left(1+x\right)\left(1-x\right)$$
$$\left(4-x^2\right)=\left(2+x\right)\left(2-x\right)$$
$$\left(9-x^2\right)=\left(3+x\right)\left(3-x\right)$$
$$Therefore,\ \left(1+x^2\right)\ \left(4-x^2\right)\ \left(9-x^2\right)>0$$
$$=>\ \left(1+x\right)\left(1-x\right)\left(2+x\right)\left(2-x\right)\left(3+x\right)\left(3-x\right)>0$$
This provides us with 3 range of x
$$1=x<-3$$
$$2=x>3$$
$$3=>x=0$$
$$if\ x<-3$$
$$\left(1-4\right)\left(1+4\right)\left(2-4\right)\left(2+4\right)\left(3-4\right)\left(3+4\right)<0$$
$$-3\cdot5\cdot-2\cdot6\cdot-1\cdot7<0$$
$$-15\cdot-12\cdot-7<0$$
$$-15\cdot-12\cdot-7<0$$
$$-1260<0$$
$$this\ does\ not\ satisfy\ the\ given\ \exp ression$$
$$if\ x>3$$
$$\left(1+4\right)\left(1-4\right)\left(2+4\right)\left(2-4\right)\left(3+4\right)\left(3-4\right)<0$$
$$5\cdot-3\cdot6\cdot-2\cdot7\cdot-1<0$$
$$-15\cdot-12\cdot-7<0$$
$$-1260<0$$
$$this\ does\ not\ satisfy\ the\ given\ inequality$$
$$if\ x=0$$
$$then\ \left(1+0\right)\left(1-0\right)\left(2+0\right)\left(2-0\right)\left(3+0\right)\left(3-0\right)>0$$
$$1\cdot1\cdot2\cdot2\cdot3\cdot3>0$$
$$36>0$$
$$this\ is\ the\ only\ integral\ value\ that\ satisfies\ the\ inequality$$
$$Since\ only\ 1\ integral\ value\ satisfies\ the\ inequality,$$
$$Answer\ =\ B$$

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8156; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

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

by Scott@TargetTestPrep » Thu Jul 02, 2020 3:48 pm

M7MBA wrote: ↑
Wed Jun 24, 2020 5:53 am
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

[spoiler]OA=B[/spoiler]

Solution:

We see that when x is 1, -1, 2, -2, 3, or -3, the value of the expression on the left hand side of the inequality will be 0, which means none of these integer values satisfy the inequality. Therefore, this leaves 0, an integer greater than 3, or an integer less than -3 as the only possible values of x. Let’s examine them.

If x = 0, we have (1)(4)(9) = 36, which is greater than 0.

If x > 3, we see that all the factors of the left hand side of the inequality will be negative. So the expression, as the product of these factors, will be also negative, which is not greater than 0.

Similarly, if x < -3, we see that all the factors of the left hand side of the inequality will be negative. So the expression, as the product of these factors, will also be negative, which is not greater than 0.

Therefore, there is only one integer value of x that satisfies the inequality, namely, 0.

Answer: B

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Founder and CEO
[email protected]

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