Difference between revisions of "1974 AHSME Problems/Problem 10"
(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...") |
|||
| Line 1: | Line 1: | ||
| − | |||
==Solution== | ==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 discriminant. Therefore, <math> (-8)^2-4(2k-1)(6)<0\implies 64-48k+24<0\implies k>\frac{11}{6} </math>. From here, we can easily see that the smallest integral value of <math> k </math> is <math> 2, \boxed{\text{B}} </math>. | 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 discriminant. Therefore, <math> (-8)^2-4(2k-1)(6)<0\implies 64-48k+24<0\implies k>\frac{11}{6} </math>. From here, we can easily see that the smallest integral value of <math> k </math> is <math> 2, \boxed{\text{B}} </math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME box|year=1974|num-b=9|num-a=11}} | ||
Revision as of 17:07, 26 May 2012
Solution
Expanding, we have 
, or 
. For this quadratic not to have real roots, it must have a negative discriminant. Therefore, 
. From here, we can easily see that the smallest integral value of 
is 
.
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 | ||
