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If 2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0, then which of the fol

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If 2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0, then which of the fol [#permalink] New post   09 Mar 2020, 03:18
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If \(2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0\), then which of the following could be the value of x ?

A. 1
B. 3
C. 5
D. 7
E. 9


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shameekv1989
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Re: If 2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0, then which of the fol [#permalink] New post   09 Mar 2020, 03:39
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If \(2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0\), then which of the following could be the value of x ?

A. 1
B. 3
C. 5
D. 7
E. 9

The product of the roots = \(e/a\) = 130/2 = 65 = 13*5 = Product of 2 co primes.

Thus one of the roots should be 5 from the given options IMO

Answer - C
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If 2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0, then which of the fol [#permalink] New post   09 Mar 2020, 05:27
Any other approach to the sum? I woild have never thought to do e/a!

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Re: If 2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0, then which of the fol [#permalink] New post   09 Mar 2020, 09:03
Bunuel wrote:
If \(2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0\), then which of the following could be the value of x ?

A. 1
B. 3
C. 5
D. 7
E. 9


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Putting x=5 in \(2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0\)

IMO C
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If 2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0, then which of the fol [#permalink] New post   10 Mar 2020, 06:10
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Bunuel wrote:
If \(2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0\), then which of the following could be the value of x ?

A. 1
B. 3
C. 5
D. 7
E. 9


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Thanks shameekv1989

due to some technical problems in which I cannot post some letters correctly to the borad, i put it in the attached file
i write down some way of thinking and related theorem on the board
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If 2x^5 − 14x^4 + 31x^3.docx [27.57 KiB]
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Re: If 2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0, then which of the fol [#permalink] New post   10 Mar 2020, 06:20
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mimishyu wrote:
Bunuel wrote:
If \(2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0\), then which of the following could be the value of x ?

A. 1
B. 3
C. 5
D. 7
E. 9


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Thanks shameekv1989

due to some technical problems in which I cannot post some letters correctly to the borad, i put it in the attached file
i write down some way of thinking and related theorem on the board


Hey mimishyu - Amazing additional information. Thanks for confirming the approach.
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Re: If 2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0, then which of the fol [#permalink] New post   10 Mar 2020, 08:01
Thank you for your appreciation,I translated all from Chinese website since I cannot find any related material in English which can better help to explain this ques, before I post I'm quite nervous whether I will do sth wrong

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If 2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0, then which of the fol [#permalink] New post   11 Mar 2020, 04:25
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If \(2x^{5}−14x^{4}+31x^{3}−64x^{2}+19x+130=0\), then which of the following could be the value of x ?

\(2x^{5}−14x^{4}+31x^{3}−64x^{2}+19x+130=\)

\((2x^{5}- 10x^{4}) -(4x^{4}- 20x^{3}) +(11x^{3} - 55x^{2}) -(9x^{2} -45x) -(26x -5*26)= \)

\(2x^{4}( x-5) -4x^{3}(x-5)+ 11x^{2}(x-5 ) -9x(x-5) -26(x-5 )=\)

\(( x-5)*(2x^{4}- 4x^{3}+ 11x^{2}-9x -26 ) =0\)

--> \(x\) could be \(5\).

Answer (C)
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Re: If 2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0, then which of the fol [#permalink] New post   30 May 2020, 06:40
2x^5−14x^4+31x^3−64x^2+19^x+130=0

product of all = -130*2= -260
Sum of all = 19/2
Now we can either go to each option and evaluate the expression
Or we could go for smart guess.

If the product is -260 then assuming that at least one integer will be there
We can go for sure that 10 will be multiple.
So from options 1 and 5 seems the right options.
Keep x = 1 and the equation won't hold.

So x = 5
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Re: If 2x^5 14x^4 + 31x^3 64x^2 + 19x + 130 = 0, then which of the fol [#permalink] New post   31 Dec 2023, 12:32
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Re: If 2x^5 14x^4 + 31x^3 64x^2 + 19x + 130 = 0, then which of the fol [#permalink]
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