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

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

by Max@Math Revolution » Tue Mar 01, 2016 11:36 pm
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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Source: — Data Sufficiency |

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by Dario@VinciaPrep » Thu Mar 03, 2016 2:03 am
1 alone:
The statement can be re-written as b=a+4, and then plugged into the equation.
The equation becomes: x^3-ax^2+ax+4x-5=0
If x is replaced with 1, the equation becomes: 1-a+a+4-5=0
This will work for any value of a.
Thus STATEMENT 1 ALONE IS SUFFICIENT.

2 alone:
If x=5 is a root of this equation, I can plug it back into the equation.
The equations becomes: 125-25a+5b-5=0
Which can be re-written as: 5a-b=24, or b=5a-24
Plugging this relationship into the initial equation, we get: x^3-ax^2+5ax-24x-5=0
Then we need to check if 1 is a root of this equation.
We get: 1-a+5a-24-5=0, which is true only if 4a=28, or a=7.
As 1 is not a root for any value of a, STATEMENT 2 ALONE IS NOT SUFFICIENT.
In order to show that, you can take two examples:
- a=7, b=11. In this case, 5 and 1 are both roots of the equation;
- a=5, b=1. In this case, 5 is a root, but 1 is not a root.

Thus the answer is (A).
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by Max@Math Revolution » Thu Mar 03, 2016 5:15 pm
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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