1974 AHSME Problems/Problem 10

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Problem

What is the smallest integral value of $k$ such that \[2x(kx-4)-x^2+6=0\] has no real roots?

$\mathrm{(A)\ } -1 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \  } 3 \qquad \mathrm{(D) \  } 4 \qquad \mathrm{(E) \  }5$

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

See Also

1974 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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