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655-705 (Hard)|   Algebra|      
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There is an equation: x^3-ax^2+bx-5=0, a and b are positive integers, 02 Mar 2016, 00:36
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62% (01:57) correct 38% (02:16) wrong based on 128 sessions

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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


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There is an equation: x^3-ax^2+bx-5=0, a and b are positive integers, 03 Mar 2016, 18:14
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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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There is an equation: x^3-ax^2+bx-5=0, a and b are positive integers, 03 Mar 2016, 21:24
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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!
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There is an equation: x^3-ax^2+bx-5=0, a and b are positive integers, 07 Mar 2016, 16:32
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sefienolte
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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-> 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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There is an equation: x^3-ax^2+bx-5=0, a and b are positive integers, 17 Nov 2022, 20:25
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MathRevolution
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.
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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
MathRevolution
sefienolte
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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\((x-5)(x^2+x+1)=0…..x^3-5x^2+x^2-5x+x-5=0………x^3-4x^2-4x-5=0\)
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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