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0 < x < 1

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0 < x < 1 [#permalink] New post 14 Oct 2017, 23:56
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Question Stats:

58% (00:50) correct 41% (01:16) wrong based on 29 sessions

0 < x < 1
Quantity A
Quantity B
\((x^3 - x)(4x + 3 )\)
\((x^2 + 1)(4x^2 + 3x)\)


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: 0 < x < 1 [#permalink] New post 16 Oct 2017, 02:01
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Doing computations we get quantity A equal to \(4x^4-4x^2+3x^3-3x\) and quantity B equal to \(4x^4+4x^2+3x^3+3x\). Now, simplifying we get A equal to \(-4x\) and B equal to \(3\). Then, whatever x between 0 and 1, B is greater!
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Re: 0 < x < 1 [#permalink] New post 15 Dec 2017, 04:14
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IlCreatore wrote:
Doing computations we get quantity A equal to \(4x^4-4x^2+3x^3-3x\) and quantity B equal to \(4x^4+4x^2+3x^3+3x\). Now, simplifying we get A equal to \(-4x\) and B equal to \(3\). Then, whatever x between 0 and 1, B is greater!


How did 4x^4+3x^3−4x^2+3x become 4x?
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Re: 0 < x < 1 [#permalink] New post 16 Dec 2017, 14:17
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Here the best approach and that works really really fast is picking numbers.

For instance, you can choose an integer 1 or -1 or a fraction but the result will be the same.

Picking -1 the first quantity will be zero and the second will be 14. Same if you pick 1.and so on and so forth.

hope this helps.

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Re: 0 < x < 1 [#permalink] New post 23 Feb 2018, 19:58
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a short cut:
extract one x from (x^3-x) we get x(x^2-1)(4x+3) on the left.
extract one x from (4x^2+3x) we get x(x^2+1)(4x+3) on the right.
x(4x+3) canceled out.
(x^2-1)<(x^2+1)
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Re: 0 < x < 1 [#permalink] New post 24 Feb 2018, 17:09
B
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Re: 0 < x < 1 [#permalink] New post 12 Mar 2018, 16:49
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correct: A (maybe includes miscalculatings!)
(x^3 - x)(4x + 3) = 4x^4 + 3x^3 - 4x^2 -3x

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

We compare these two and simplify as much as possible.
4x^4 + 3x^3 - 4x^2 -3x vs 4x^4 + 3x^3 + 4x^2 + 3x
Omitting 4x^4 + 3x^3:
- 4x^2 -3x vs 4x^2 + 3x
-(4x^2 + 3x) vs (4x^2 + 3x)
a) When 4x^2 + 3x is negative then the value in the left will be positive when multiplied by it’s negative sign. And the right value will be negative and thus less.
4x^2 + 3x < 0 —> 4x^2 < -3x —> x < -3/4
b) When 4x^2 + 3x is positive the value in the right will be bigger.
4x^2 + 3x > 0 —> 4x^2 > -3x —> x > -3/4

BUT
it's mentioned that x is between 0 and 1 and it will be case b. so A is bigger than B

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Re: 0 < x < 1 [#permalink] New post 26 Feb 2019, 07:27
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Hello Sandy,

x has no specific value. x can be positive or negative or zero. If so, why the answer is not D since (A) -4x^2-3x and B) 4x^2 + 3x have same value if x=0?
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Re: 0 < x < 1 [#permalink] New post 26 Feb 2019, 07:44
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Carcass wrote:


0 < x < 1

Quantity A
Quantity B
\((x^3 - x)(4x + 3)\)
\((x^2 + 1)(4x^2 + 3x)\)


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.



We can solve this question using matching operations

Given:
Quantity A: (x³ - x)(4x + 3)
Quantity B: (x² + 1)(4x² + 3x)

Expand both sides to get:
Quantity A: 4x⁴ + 3x³ - 4x² - 3x
Quantity B: 4x⁴ + 3x³ + 4x² + 3x

Subtract 4x⁴ from both sides to get:
Quantity A: 3x³ - 4x² - 3x
Quantity B: 3x³ + 4x² + 3x

Subtract 3x³ from both sides to get:
Quantity A: -4x² - 3x
Quantity B: 4x² + 3x

Add 4x² to both sides to get:
Quantity A: -3x
Quantity B: 8x² + 3x

Add 3x to both sides to get:
Quantity A: 0
Quantity B: 8x² + 6x

Divide both sides by 2 to get:
Quantity A: 0
Quantity B: 4x² + 3x

Factor to get:
Quantity A: 0
Quantity B: x(4x + 3)


GIVEN: 0 < x < 1
This means x is POSITIVE, and it means (4x + 3) is POSITIVE

So, we have:
Factor to get:
Quantity A: 0
Quantity B: POSITIVE (POSITIVE) = POSITIVE

Answer: B

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Re: 0 < x < 1 [#permalink] New post 26 Feb 2019, 07:46
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JelalHossain wrote:
Hello Sandy,

x has no specific value. x can be positive or negative or zero. If so, why the answer is not D since (A) -4x^2-3x and B) 4x^2 + 3x have same value if x=0?


The OP posted 0 < x < 1 in the subject line, but forgot to add it to the question.
I have since added that information to the question.

Cheers,
Brent
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Re: 0 < x < 1 [#permalink] New post 26 Feb 2019, 07:58
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Thank you Sir for adding that x is between the hammer and the anvil.

Regards
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Re: 0 < x < 1   [#permalink] 26 Feb 2019, 07:58
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