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: 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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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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\(x^3-ax^2+bx-5=0\)
Let us substitute x=1, to find the condition that should be met.

\(x^3-ax^2+bx-5=0\)
\(1^3-a*1^2+b*1-5=0\)
\(1-a+b-5=0\)
\(b-a=4\)

1) b-a=4
Meets the condition.
Sufficient

2)x=5 is a root of this equation
We know the product of three roots is -(-5)/1 or 5.
Thus, the roots could be -1*-1*5 or 1*1*5.
But, taking roots -1, -1 and 5 will make b negative. Hence, only possibility is 1, 1 and 5 and the cubic equation is \(x^3-7x^2+11x-5=0\)
Sufficient


MathRevolution, you are wrong in the OA and also in the below reply


\((x-5)(x^2+x+1)=0…..x^3-5x^2+x^2-5x+x-5=0………x^3-4x^2-4x-5=0\)
So a=4 and b=-4 but b is supposed to be a positive integer.
So, the example is not valid.
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