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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## If f(x)=ax^2+bx+c, where a, b and c are integers, is b=0? tagged by: Max@Math Revolution ##### This topic has 2 expert replies and 0 member replies ### GMAT/MBA Expert ## If f(x)=ax^2+bx+c, where a, b and c are integers, is b=0? ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult [GMAT math practice question] If f(x)=ax^2+bx+c, where a, b and c are integers, is b=0? 1) f(49)=f(-49)=0 2) f(0)f(49)=0 _________________ Math Revolution Finish GMAT Quant Section with 10 minutes to spare. The one-and-only Worldâ€™s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. Only$149 for 3 month Online Course
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Max@Math Revolution wrote:
[GMAT math practice question]

If f(x)=ax^2+bx+c, where a, b and c are integers, is b=0?

1) f(49)=f(-49)=0
2) f(0)f(49)=0
Statement 1:
Since (49, 0) and (-49, 0) are both solutions, the quadratic must be as follows:
f(x) = (x-49)(x+49) = xÂ² - 49Â²
Thus, the answer to the question stem is YES.
SUFFICIENT.

Statement 2:
If x-49 is a factor of the equation, then f(49) = 0.
Case 1: f(x) = (x-49)(x+49) = xÂ² - 49Â², with the result that f(0)f(49) = f(0) * 0 = 0
In this case, b=0, so the answer to question stem is YES.
Case 2: f(x) = (x-49)(x+1) = xÂ² - 48x - 49, with the result that f(0)f(49) = f(0) * 0 = 0
In this case, b=-48, so the answer to the question stem is NO.
INSUFFICIENT.

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

In order to have b =0, f(x) must be symmetric about the y-axis. Thus, condition 1) is sufficient.
By the factor theorem, condition 1) tells us that f(x) = a(x-49)(x+49) = a(x^2 - 49^2) = ax^2 - 49^2a and b = 0.
It is sufficient.

Condition 2)
If f(0)f(49) = 0 then either f(0)= 0 or f(49) = 0.
Note that f(0) = c.
If a = 1, b = 0 and c = 0, then f(0) = 0, so f(0)f(49) = 0, and the answer is â€˜yesâ€™.
If a = a, b = 1 and c = 0, then f(0) = 0, so f(0)f(49) = 0, and the answer is â€˜noâ€™.
Thus, condition 2) is not sufficient, since it does not yield a unique solution.

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