1974 AHSME Problems/Problem 10

Revision as of 18:06, 26 May 2012 by Admin25 (talk | contribs) (Created page with " ==Solution== Expanding, we have <math> 2kx^2-8x-x^2+6=0 </math>, or <math> (2k-1)x^2-8x+6=0 </math>. For this quadratic not to have real roots, it must have a negative discrimi...")
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Solution

Expanding, we have $2kx^2-8x-x^2+6=0$, or $(2k-1)x^2-8x+6=0$. For this quadratic not to have real roots, it must have a negative discriminant. Therefore, $(-8)^2-4(2k-1)(6)<0\implies 64-48k+24<0\implies k>\frac{11}{6}$. From here, we can easily see that the smallest integral value of $k$ is $2, \boxed{\text{B}}$.