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There is an equation: x^3-ax^2+bx-5=0, a and b are positive integers,

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There is an equation: x^3-ax^2+bx-5=0, a and b are positive integers, [#permalink]

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There is an equation: x^3-ax^2+bx-5=0, a and b are positive integers, is x=1 a root of this equation?

1) b-a=4
2)x=5 is a root of this equation


* A solution will be posted in two days.
[Reveal] Spoiler: OA

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Re: There is an equation: x^3-ax^2+bx-5=0, a and b are positive integers, [#permalink]

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New post 03 Mar 2016, 18:14
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

There is an equation: x3-ax2+bx-5=0, a and b are positive integers, is x=1 a root of this equation?

1) b-a=4
2) x=5 is a root of this equation


Modify the original condition and the question. Substitute x=1 and 1-a+b-5=0? is derived. That is, it becomes b-a=4?, which is yes and sufficient.
Thus, A is the answer.
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Re: There is an equation: x^3-ax^2+bx-5=0, a and b are positive integers, [#permalink]

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New post 03 Mar 2016, 21:24
Can you explain why the second statement is not sufficient? I guess what you mean is that knowing a second root to the problem doesn't tell us much about whether x=1 is a root? Thanks!

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Re: There is an equation: x^3-ax^2+bx-5=0, a and b are positive integers, [#permalink]

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sefienolte wrote:
Can you explain why the second statement is not sufficient? I guess what you mean is that knowing a second root to the problem doesn't tell us much about whether x=1 is a root? Thanks!



-> Since this is cubic function, even if x=5 is an answer, you cannot figure out if x=1 is an answer as (x-5)(x^2+x+1)=0 is possible as well.
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Re: There is an equation: x^3-ax^2+bx-5=0, a and b are positive integers, [#permalink]

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Re: There is an equation: x^3-ax^2+bx-5=0, a and b are positive integers,   [#permalink] 21 Oct 2017, 13:39
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