Difference between revisions of "2019 AMC 8 Problems/Problem 20"
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==Solution 1== | ==Solution 1== | ||
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 Equation|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 Equation|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. | ||
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+ | Note: <math>x^2</math> can be two possible things and still be the same, and in <math>(x^2+4)^2</math>, <math>x^2+4</math> can also be negative or positive, so <math>2 \cdot 2 = 4</math>. This is a very fast solution. | ||
==Solution 2== | ==Solution 2== |
Latest revision as of 00:22, 7 June 2021
Contents
Problem 20
How many different real numbers satisfy the equation
Solution 1
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.
Note: can be two possible things and still be the same, and in , can also be negative or positive, so . This is a very fast solution.
Solution 2
We can expand to get , so now our equation is . Subtracting from both sides gives us . Now, we can factor the left hand side to get . If and/or equals , then the whole left side will equal . Since the solutions can be both positive and negative, we have solutions: (we can find these solutions by setting and equal to and solving for ). So the answer is .
~UnstoppableGoddess
Solution 3
Associated Video - https://www.youtube.com/watch?v=Q5yfodutpsw
Solution 4
https://youtu.be/5BXh0JY4klM (Uses a difference of squares & factoring method, different from above solutions)
Solution 5 (Fast)
https://www.youtube.com/watch?v=44vrsk_CbF8&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=2 ~ MathEx
Video Solution
-Happytwin
Video Solution
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=eXJnG96Xqp4&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=21
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.