1974 AHSME Problems/Problem 10

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 $2, \boxed{\text{B}}$ .

1974 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1974 AHSME Problems/Problem 10

Solution

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