2017 AIME II Problems/Problem 7: Difference between revisions
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==Solution== | ==Solution== | ||
<math>\boxed{501}</math> | <math>\boxed{501}</math> | ||
kx=(x+2)^2 | |||
x^2+(4-k)x+4=0 ...(1) | |||
the equation has solution | |||
D=(4-k)^2-16=k(k-8)>=0 | |||
so k<=0 or k>=8 | |||
becuase k can't be zero or the original equation will be meanless. | |||
there are 3 cases | |||
1:k=8 | |||
then x=2, which is satisified the question. | |||
2:k<0 | |||
then one solution of the equation(1) should be in (-2,0) and another is out of it or the origin equation will be meanless. | |||
then we get 2 inequities | |||
-2<( k-4 + sqrt( k(k-8) ) )/2<0 | |||
( k-4 - sqrt( k(k-8) ) )/2<-2 | |||
notice k<0<sqrt(k(k-8)) and (4-k)^2=k(k-8)+16>k(k-8) | |||
we know in this case, there is always and only one solution for the orign equation. | |||
3:k>8 | |||
similar to case2 we can get inequity | |||
( k-4 - sqrt( k(k-8) ) )/2 < 0 <( k-4 + sqrt( k(k-8) ))/2 | |||
and there are always 2 solution for the origin equation, so this case is not satisfied. | |||
so we get k<0 or k=8 | |||
because k belong to [-500,500], the answer is 501 | |||
=See Also= | =See Also= | ||
{{AIME box|year=2017|n=II|num-b=6|num-a=8}} | {{AIME box|year=2017|n=II|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:33, 23 March 2017
Problem
Find the number of integer values of 
in the closed interval ![$[-500,500]$](http://latex.artofproblemsolving.com/7/6/8/768f4e11a324ee335c6a0689a310fdaa8c33c921.png)
for which the equation 
has exactly one real solution.
Solution
kx=(x+2)^2 x^2+(4-k)x+4=0 ...(1) the equation has solution D=(4-k)^2-16=k(k-8)>=0 so k<=0 or k>=8 becuase k can't be zero or the original equation will be meanless. there are 3 cases
1:k=8 then x=2, which is satisified the question.
2:k<0 then one solution of the equation(1) should be in (-2,0) and another is out of it or the origin equation will be meanless. then we get 2 inequities -2<( k-4 + sqrt( k(k-8) ) )/2<0 ( k-4 - sqrt( k(k-8) ) )/2<-2 notice k<0<sqrt(k(k-8)) and (4-k)^2=k(k-8)+16>k(k-8) we know in this case, there is always and only one solution for the orign equation.
3:k>8 similar to case2 we can get inequity ( k-4 - sqrt( k(k-8) ) )/2 < 0 <( k-4 + sqrt( k(k-8) ))/2 and there are always 2 solution for the origin equation, so this case is not satisfied.
so we get k<0 or k=8
because k belong to [-500,500], the answer is 501
See Also
| 2017 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America. 
