In the equation x^2−7x+n=0, n is a constant and x is a variable. Is (x−6) a factor of x^2−7x+n=0?
(1) n=6
(2) (x−1) is a factor of x2−7x+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^2−k·x+24=0, is x=−6?
(1) (x+6) is a factor of x2-k·x+24, where k is a constant, and x is a variable.
(2) (x+4) is a factor of x2-k·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 (e·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 (e·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!
(1) n=6
(2) (x−1) is a factor of x2−7x+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^2−k·x+24=0, is x=−6?
(1) (x+6) is a factor of x2-k·x+24, where k is a constant, and x is a variable.
(2) (x+4) is a factor of x2-k·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 (e·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 (e·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!
