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Difference between revisions of "1974 AHSME Problems/Problem 10"
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==Problem==
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>What is the smallest integral value of <math> k </math> such that
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<cmath> 2x(kx-4)-x^2+6=0 </cmath>
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has no real roots?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
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<math> \mathrm{(A)\ } -1 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \  } 3 \qquad \mathrm{(D) \  } 4 \qquad \mathrm{(E) \  }5  </math>
  
 
==Solution==
 
==Solution==
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==See Also==
 
==See Also==
 
{{AHSME box|year=1974|num-b=9|num-a=11}}
 
{{AHSME box|year=1974|num-b=9|num-a=11}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 16:14, 18 November 2015

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 (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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