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by uniyal01 » Thu Jun 23, 2016 2:34 pm
In the equation x^2âˆ’7x+n=0, n is a constant and x is a variable. Is (xâˆ’6) a factor of x^2âˆ’7x+n=0?

(1) n=6

(2) (xâˆ’1) is a factor of x2âˆ’7x+n=0.

Ans:
D

But for 1) if we put n=6 into the equation wouldn't that yield 2 different values for x? namely x=6 or x=1? So why's this statement sufficient?

This line of logic is consistent with a a similar question,

If x^2âˆ’kÂ·x+24=0, is x=âˆ’6?

(1) (x+6) is a factor of x2-kÂ·x+24, where k is a constant, and x is a variable.

(2) (x+4) is a factor of x2-kÂ·x+24, where k is a constant, and x is a variable.

Ans=E because Stat.(1) means either e or f are 6, so the other must be 4 (eÂ·f=24). The roots therefore are x1=-6 x2=-4, which means you cannot determine whether x is -6. Hence, Stat.(1)->Maybe->IS->BCE.

Stat.(2) means either e or f are 4, so the other must be 6 (eÂ·f=24). The roots therefore are x1=-4 x2=-6, which means you cannot determine whether x is -6. Hence, Stat.(2)->Maybe->IS->CE.

Stat.(1+2) mean the factored form is (x+4)(x+6), so either e or f is 4, and the other is 6. The roots therefore are x1=-4 x2=-6, which means you cannot determine whether x is -6. Hence, Stat.(1+2)->Maybe->IS->E.

Thank you!

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by Matt@VeritasPrep » Thu Jun 23, 2016 3:46 pm
You would get two different values, but the problem isn't asking for the value of x, it's asking (in a clumsy way) whether 6 is one possible value of x.

S1 tells us that

xÂ² - 7x + 6 = 0

Plugging in 6, we find 6Â² - 42 + 6 does = 0, so x = 6 is one solution; SUFFICIENT.

S2 tells us that

x = 1 is one solution, so

1Â² - 7*1 + n = 0, and n = 6. This is the same as the first statement; SUFFICIENT.

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by Matt@VeritasPrep » Thu Jun 23, 2016 3:47 pm
Also, be sure to post DS questions in the DS forum, not the PS forum - that will help us answer them more quickly.

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