Difference between revisions of "1974 AHSME Problems/Problem 10"

Revision as of 11:43, 5 July 2013

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

1974 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 {{AHSME box|year=1974|num-b=9|num-a=11}}
 [[Category:Introductory Algebra Problems]]
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Difference between revisions of "1974 AHSME Problems/Problem 10"

Revision as of 11:43, 5 July 2013

Problem

Solution

See Also