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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
Are You Up For the Challenge: 700 Level Questions
A. 1
B. 3
C. 5
D. 7
E. 9
Are You Up For the Challenge: 700 Level Questions
shameekv1989
GMAT 3: 720 Q50 V38
Posts: 816
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
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
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)
\(2x^{5}−14x^{4}+31x^{3}−64x^{2}+19x+130=\)
\((2x^{5}- 10x^{4}) -(4x^{4}- 20x^{3}) +(11x^{3} - 55x^{2}) -(9x^{2} -45x) -(26x -5*26)= \)
\(2x^{4}( x-5) -4x^{3}(x-5)+ 11x^{2}(x-5 ) -9x(x-5) -26(x-5 )=\)
\(( x-5)*(2x^{4}- 4x^{3}+ 11x^{2}-9x -26 ) =0\)
--> \(x\) could be \(5\).
Answer (C)
