Difference between revisions of "2019 AMC 8 Problems/Problem 20"
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We can expand <math>(x^2-5)^2</math> to get <math>x^4-10x^2+25</math>, so now our equation is <math>x^4-10x^2+25=16</math>. Subtracting <math>16</math> from both sides gives us <math>x^4-10x^2+9=0</math>. Now, we can factor the left hand side to get <math>(x^2-9)(x^2-1)=0</math>. If <math>x^2-9</math> and/or <math>x^2-1</math> equals <math>0</math>, then the whole left side will equal <math>0</math>. Since the solutions can be both positive and negative, we have <math>4</math> solutions: <math>-3,3,-1,1</math> (we can find these solutions by setting <math>x^2-9</math> and <math>x^2-1</math> equal to <math>0</math> and solving for <math>x</math>). So, the answer is <math>\boxed{\textbf{(D) }4}</math>. | We can expand <math>(x^2-5)^2</math> to get <math>x^4-10x^2+25</math>, so now our equation is <math>x^4-10x^2+25=16</math>. Subtracting <math>16</math> from both sides gives us <math>x^4-10x^2+9=0</math>. Now, we can factor the left hand side to get <math>(x^2-9)(x^2-1)=0</math>. If <math>x^2-9</math> and/or <math>x^2-1</math> equals <math>0</math>, then the whole left side will equal <math>0</math>. Since the solutions can be both positive and negative, we have <math>4</math> solutions: <math>-3,3,-1,1</math> (we can find these solutions by setting <math>x^2-9</math> and <math>x^2-1</math> equal to <math>0</math> and solving for <math>x</math>). So, the answer is <math>\boxed{\textbf{(D) }4}</math>. | ||
− | ~UnstoppableGoddess | + | ~UnstoppableGoddess |
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | |||
+ | Subtract 16 from both sides and factor using difference of squares: | ||
+ | |||
+ | |||
+ | |||
+ | <cmath>(x^2 - 5)^2 = 16 </cmath> | ||
+ | <cmath>(x^2 - 5)^2 - 16 =0 </cmath> | ||
+ | <cmath>(x^2 - 5)^2 - 4^2 = 0 </cmath> | ||
+ | <cmath>[(x^2 - 5)-4][(x^2 - 5) + 4] = 0</cmath> | ||
+ | <cmath>(x^2 - 9)(x^2 - 1) =0 </cmath> | ||
+ | <cmath>(x+3)(x-3)(x+1)(x-1) = 0 </cmath> | ||
+ | |||
+ | |||
+ | Quite obviously, this equation has <math>\boxed{\textbf{(D) }4}</math> solutions. | ||
==Solution 3== | ==Solution 3== |
Revision as of 23:16, 10 January 2023
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.
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
Subtract 16 from both sides and factor using difference of squares:
Quite obviously, this equation has solutions.
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 (video of Solution 1)
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://youtu.be/x4cF3o3Fzj8
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.