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For how many integer values of x, is 3x3+2x+8<15?
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Updated on: 13 Feb 2017, 07:20
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For how many integer values of x, is \(3x3+2x+8<15\)? A. 2 B. 3 C. 4 D. 5 E. 6 Attachment:
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Originally posted by hazelnut on 13 Feb 2017, 06:42.
Last edited by hazelnut on 13 Feb 2017, 07:20, edited 1 time in total.




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Re: For how many integer values of x, is 3x3+2x+8<15?
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13 Feb 2017, 07:08
we can solve this by following ways 
1. First take the positive parts of mod values 
3x3 = 3x3 2x+8 = 2x+8
So, it boils down to 
3x3+2x+8 < 15 x<2
2. Now consider the negative parts of them 
3x3 = 3x+3 2x+8 = 2x8
3x+32x8<15 x<4 x>4
So , we can conclude that: 4<x<2 Thus, we will have  5 integers, which will satisfy this equation.




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Re: For how many integer values of x, is 3x3+2x+8<15?
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13 Feb 2017, 06:59



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Re: For how many integer values of x, is 3x3+2x+8<15?
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13 Feb 2017, 07:11
ziyuenlau wrote: How to resolve this question?
For how many integer values of x, is \(3x3+2x+8<15\)? ziyuenlauPlease provide answer choices. Answer must be 5 ...please confirm



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Re: For how many integer values of x, is 3x3+2x+8<15?
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13 Feb 2017, 08:16
\(3x3 = 0\) then \(x = 1\) \(2x+8 = 0\) then \(x = 4\) there can be 3 range \(x\leq{4}\)  \(4 < x < 1\)  \(x\geq{1}\) 1. When \(x\leq{4}\), then both \(3x3\) and \(2x+8\) will be negative. \(3x+3 2x8 < 15\) \(5x5 < 15\) \(x > 4\) (this is opposite to \(x\leq{4}\). ) Hence \(x\leq{4}\) is not a possibility. 2. When \(4 < x < 1\), then \(3x3\) would still be negative but \(2x+8\) will be positive. hence \(3x+3 +2x+8 < 15\) x+11 < 15 \(x < 4\) i.e. \(x > 4\) (this lies within \(4 < x < 1\)) so correct range. 3. When \(x\geq{1}\), then both \(3x3\) and \(2x+8\) will be positive. \(3x3+2x+8 < 15\) \(5x+5 < 15\) \(x+1 < 3\) \(x < 2\) this range is also fine because \(x\geq{1}\). so x must be 1. So our range is \(4 <\)\(x\geq{1}\), so howmany integers within this range? 3, 2, 1, 0 and 1 Count is 5 Answer D.
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Re: For how many integer values of x, is 3x3+2x+8<15?
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13 Feb 2017, 12:27
ziyuenlau wrote: For how many integer values of x, is \(3x3+2x+8<15\)?
A. 2 B. 3 C. 4 D. 5 E. 6 \(3x3+2x+8<15\) \(3*x1+2*x+4<15\) We want the values of x such that the sum of "thrice their distance from 1" and "twice their distance from 4" is less than 15. Let's try to find the point where this distance is equal to 15. ........................ (4) ...................................... (0) ........... (1) .......................... The distance between 4 and 1 is 5. Thrice this distance is 15. So at the point x = 4, the sum will be 15. As we move to the right of 4, the sum will reduce (since the twice component will keep increasing). At x = 1, the sum becomes 0 + 2*5 = 10. What happens when you go to the right of 1? Now the sum starts increasing since the thrice components increasing now. At x = 2, the sum becomes 3*1 + 2*6 = 15. To the right of 2, the sum will keep increasing. So the sum will be less than 15 between 4 and 2. This gives us 5 integer values (3, 2, 1, 0, 1). Answer (D)
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Re: For how many integer values of x, is 3x3+2x+8<15?
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01 Nov 2017, 19:00
Bunuel wrote: Are there any more questions apart from the mentioned above on the same topic?



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Re: For how many integer values of x, is 3x3+2x+8<15?
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01 Nov 2017, 21:09
Buttercup3 wrote: Bunuel wrote: Are there any more questions apart from the mentioned above on the same topic? 9. Inequalities For more check Ultimate GMAT Quantitative Megathread Hope it helps.
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For how many integer values of x, is 3x3+2x+8<15?
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27 Dec 2017, 01:30
VeritasPrepKarishma wrote: ziyuenlau wrote: For how many integer values of x, is \(3x3+2x+8<15\)?
A. 2 B. 3 C. 4 D. 5 E. 6 \(3x3+2x+8<15\) \(3*x1+2*x+4<15\) We want the values of x such that the sum of "thrice their distance from 1" and "twice their distance from 4" is less than 15. Let's try to find the point where this distance is equal to 15. ........................ (4) ...................................... (0) ........... (1) .......................... The distance between 4 and 1 is 5. Thrice this distance is 15. So at the point x = 4, the sum will be 15. As we move to the right of 4, the sum will reduce (since the twice component will keep increasing). At x = 1, the sum becomes 0 + 2*5 = 10. What happens when you go to the right of 1? Now the sum starts increasing since the thrice components increasing now. At x = 2, the sum becomes 3*1 + 2*6 = 15. To the right of 2, the sum will keep increasing. So the sum will be less than 15 between 4 and 2. This gives us 5 integer values (3, 2, 1, 0, 1). Answer (D) Mam, my question in when x < = 4, then for if x = 5 , 2x+8 is negative. but we are also using x = 4, then 2x+8 becomes "0". then why we are not considering 2x+8 = 0 for x = 4. I don't know whether I made myself clear to you or not
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Re: For how many integer values of x, is 3x3+2x+8<15?
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06 Jan 2019, 06:33
HKD1710 wrote: \(3x3 = 0\) then \(x = 1\) \(2x+8 = 0\) then \(x = 4\)
there can be 3 range \(x\leq{4}\)  \(4 < x < 1\)  \(x\geq{1}\)
1. When \(x\leq{4}\), then both \(3x3\) and \(2x+8\) will be negative.
\(3x+3 2x8 < 15\)
\(5x5 < 15\)
\(x > 4\) (this is opposite to \(x\leq{4}\). )
Hence \(x\leq{4}\) is not a possibility.
2. When \(4 < x < 1\), then \(3x3\) would still be negative but \(2x+8\) will be positive. hence
\(3x+3 +2x+8 < 15\)
x+11 < 15
\(x < 4\) i.e. \(x > 4\) (this lies within \(4 < x < 1\)) so correct range.
3. When \(x\geq{1}\), then both \(3x3\) and \(2x+8\) will be positive.
\(3x3+2x+8 < 15\)
\(5x+5 < 15\)
\(x+1 < 3\)
\(x < 2\) this range is also fine because \(x\geq{1}\). so x must be 1.
So our range is \(4 <\)\(x\geq{1}\), so howmany integers within this range?
3, 2, 1, 0 and 1
Count is 5
Answer D. but took 3.5 mins, i'd expect quicker approach



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Re: For how many integer values of x, is 3x3+2x+8<15?
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25 Nov 2019, 06:11
hazelnut wrote: For how many integer values of x, is \(3x3+2x+8<15\)?
A. 2 B. 3 C. 4 D. 5 E. 6 \(positive:3x3≥0…3x≥3…x≥1…negative:x<1\) \(positive:2x+8≥0…2x≥8…x≥4…negative:x<4\) \(range:(neg)(4)(pos,neg)(1)(pos)\) \(x≥1:3x3+2x+8<15…3x3+2x+8<15…5x<10…x<2:1≤x<2=[1]\) \(4≤x<1:…3x+3+2x+8<15…x<4…x>4:4<x<1=[3,2,1,0]\) \(x<4:…3x+32x8<15…5x<20…x>4:invalid=x<4\) \(x=[3,2,1,0,1]=5\) Ans (D)



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Re: For how many integer values of x, is 3x3+2x+8<15?
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12 Dec 2019, 09:16
faltanfaltan wrote: HKD1710 wrote: \(3x3 = 0\) then \(x = 1\) \(2x+8 = 0\) then \(x = 4\)
there can be 3 range \(x\leq{4}\)  \(4 < x < 1\)  \(x\geq{1}\)
1. When \(x\leq{4}\), then both \(3x3\) and \(2x+8\) will be negative.
\(3x+3 2x8 < 15\)
\(5x5 < 15\)
\(x > 4\) (this is opposite to \(x\leq{4}\). )
Hence \(x\leq{4}\) is not a possibility.
2. When \(4 < x < 1\), then \(3x3\) would still be negative but \(2x+8\) will be positive. hence
\(3x+3 +2x+8 < 15\)
x+11 < 15
\(x < 4\) i.e. \(x > 4\) (this lies within \(4 < x < 1\)) so correct range.
3. When \(x\geq{1}\), then both \(3x3\) and \(2x+8\) will be positive.
\(3x3+2x+8 < 15\)
\(5x+5 < 15\)
\(x+1 < 3\)
\(x < 2\) this range is also fine because \(x\geq{1}\). so x must be 1.
So our range is \(4 <\)\(x\geq{1}\), so howmany integers within this range?
3, 2, 1, 0 and 1
Count is 5
Answer D. but took 3.5 mins, i'd expect quicker approach Did you came up with one?




Re: For how many integer values of x, is 3x3+2x+8<15?
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