(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!