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What is the product of the factors of twice the sum of the roots of th

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What is the product of the factors of twice the sum of the roots of th  [#permalink]

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New post 10 Dec 2018, 00:49
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

53% (01:24) correct 47% (01:24) wrong based on 72 sessions

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Re: What is the product of the factors of twice the sum of the roots of th  [#permalink]

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New post 10 Dec 2018, 00:56
Sum of roots of the equation is \(2x^2 - 4x - 6 = 0\) is 2. Hence twice the sum of roots is 4.

Using, \(ax^2 +bx +c = 0\) has sum of roots -b/a

Now, factors of 4 are = 1, 2, 4. The product of these is 8.

Hence Option (C) is our choice.

Best,
Gladi

Bunuel wrote:
What is the product of the factors of twice the sum of the roots of the equation \(2x^2 - 4x - 6 = 0\)?

A. 2
B. 4
C. 8
D. 16
E. 32

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Gladi



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Re: What is the product of the factors of twice the sum of the roots of th  [#permalink]

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New post 10 Dec 2018, 04:05
Bunuel wrote:
What is the product of the factors of twice the sum of the roots of the equation \(2x^2 - 4x - 6 = 0\)?

A. 2
B. 4
C. 8
D. 16
E. 32



\(2x^2 - 4x - 6 = 0\)

(x-3)(x+1)
x=3 and -1

2(3-1)= 4
factors of 4= 1,2,4 its product 1*2*4 = 8 IMO C
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What is the product of the factors of twice the sum of the roots of th  [#permalink]

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New post 10 Dec 2018, 09:15
First, we need to find the roots of the given equation.

\(2x^2\)-4x-6 = 0
Roots of this equation are 3 and -1

Second, we need to find twice the sum of the roots of the equation.
Sum = 3+(-1) = 2 and twice the sum will be 4

Third, factors of 4 are 1,2 & 4

Product of the factors = \(1*2*4\) = 8

IMO C is our answer
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What is the product of the factors of twice the sum of the roots of th   [#permalink] 10 Dec 2018, 09:15
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