russ9 wrote:
Bunuel wrote:
If r and s are the roots of the equation x^2+bx+c=0, where b and c are constants, is rs<0?
Viete's theorem states that for the rootsx_1 andx_2 of a quadratic equationax^2+bx+c=0:
x_1+x_2=\frac{-b}{a} ANDx_1*x_2=\frac{c}{a} .
Thus according to the abovers=\frac{c}{1}=c . So, we are basically asked whetherc<0 .
(1) b<0. Not sufficient.
(2) c<0. Directly answers the question. Sufficient.
Answer: B.
Hope it helps.
Viete's theorem states that for the rootsx_1 andx_2 of a quadratic equationax^2+bx+c=0:
x_1+x_2=\frac{-b}{a} ANDx_1*x_2=\frac{c}{a} .
Thus according to the abovers=\frac{c}{1}=c . So, we are basically asked whetherc<0 .
(1) b<0. Not sufficient.
(2) c<0. Directly answers the question. Sufficient.
Answer: B.
Hope it helps.
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