Difference between revisions of "2019 AMC 8 Problems/Problem 20"
m (→Solution: Fixed the letter) |
Aopsuser101 (talk | contribs) (→Solution) |
||
Line 6: | Line 6: | ||
==Solution== | ==Solution== | ||
We have that <math>(x^2-5)^2 = 16</math> if and only if <math>x^2-5 = \pm 4</math>. If <math>x^2-5 = 4</math>, then <math>x^2 = 9 \implies x = \pm 3</math>, giving 2 solutions. If <math>x^2-5 = -4</math>, then <math>x^2 = 1 \implies x = \pm 1</math>, giving 2 more solutions. All four of these solutions work, so the answer is <math>\boxed{\textbf{(D)} 4}</math>. Further, the equation is a quartic in <math>x</math>, so by the [https://artofproblemsolving.com/wiki/index.php/Fundamental_Theorem_of_Algebra fundamental theorem of algebra], there can be at most four real solutions. | We have that <math>(x^2-5)^2 = 16</math> if and only if <math>x^2-5 = \pm 4</math>. If <math>x^2-5 = 4</math>, then <math>x^2 = 9 \implies x = \pm 3</math>, giving 2 solutions. If <math>x^2-5 = -4</math>, then <math>x^2 = 1 \implies x = \pm 1</math>, giving 2 more solutions. All four of these solutions work, so the answer is <math>\boxed{\textbf{(D)} 4}</math>. Further, the equation is a quartic in <math>x</math>, so by the [https://artofproblemsolving.com/wiki/index.php/Fundamental_Theorem_of_Algebra fundamental theorem of algebra], there can be at most four real solutions. | ||
+ | |||
+ | |||
+ | Side-note. | ||
+ | Did the problem writers run out of ideas? Looks like it. | ||
==See Also== | ==See Also== |
Revision as of 21:26, 22 November 2019
Problem 20
How many different real numbers satisfy the equation
Solution
We have that if and only if
. If
, then
, giving 2 solutions. If
, then
, giving 2 more solutions. All four of these solutions work, so the answer is
. Further, the equation is a quartic in
, so by the fundamental theorem of algebra, there can be at most four real solutions.
Side-note.
Did the problem writers run out of ideas? Looks like it.
See Also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.