2002 AMC 10B Problems/Problem 13
Contents
Problem
Find the value(s) of such that
is true for all values of
.
Solution
We have .
As must be true for all
, we must have
, hence
.
Solution 2
Since we want only the -variable to be present, we move the terms only with the
-variable to one side, thus constructing
to
. For there to be infinite solutions for
and there is no
, we simply find a value of
such that the equation is symmetrical. Therefore,
There is only one solution, namely $x = \box{\dfrac{3}{2}}$ (Error compiling LaTeX. Unknown error_msg)
See Also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.