Difference between revisions of "2000 AMC 10 Problems/Problem 24"
5849206328x (talk | contribs) (New page: ==Problem== ==Solution== ==See Also== {{AMC10 box|year=2000|num-b=23|num-a=25}}) |
|||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| + | Let <math>f</math> be a function for which <math>f\left(\frac{x}{3}\right)=x^2+x+1</math>. Find the sum of all values of <math>z</math> for which <math>f(3z)=7</math>. | ||
| + | |||
| + | <math>\mathrm{(A)}\ -\frac{1}{3} \qquad\mathrm{(B)}\ -\frac{1}{9} \qquad\mathrm{(C)}\ 0 \qquad\mathrm{(D)}\ \frac{5}{9} \qquad\mathrm{(E)}\ \frac{5}{3}</math> | ||
==Solution== | ==Solution== | ||
| + | |||
| + | In the definition of <math>f</math>, let <math>x=9z</math>. We get: <math>f(3z)=(9z)^2+(9z)+1</math>. As we have <math>f(3z)=7</math>, we must have <math>f(3z)-7=0</math>, in other words <math>81z^2 + 9z - 6 = 0</math>. | ||
| + | |||
| + | One can now either explicitly compute the roots, or use [[Vieta's formulas]]. According to them, the sum of the roots of <math>ax^2+bx+c=0</math> is <math>-\frac ba</math>. In our case this is <math>-\frac{9}{81}=\boxed{-\frac 19}</math>. | ||
| + | |||
| + | (Note that for the above approach to be completely correct, we should additionally verify that there actually are two distinct real roots. This is, for example, obvious from the facts that <math>81>0</math> and <math>f(0)-7 < 0</math>.) | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2000|num-b=23|num-a=25}} | {{AMC10 box|year=2000|num-b=23|num-a=25}} | ||
Revision as of 08:53, 11 January 2009
Problem
Let 
be a function for which 
. Find the sum of all values of 
for which 
.
Solution
In the definition of , let 
. We get: 
. As we have , we must have 
, in other words 
.
One can now either explicitly compute the roots, or use Vieta's formulas. According to them, the sum of the roots of 
is 
. In our case this is 
.
(Note that for the above approach to be completely correct, we should additionally verify that there actually are two distinct real roots. This is, for example, obvious from the facts that 
and 
.)
See Also
| 2000 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
