Difference between revisions of "2019 AMC 8 Problems/Problem 20"
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<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8</math> | <math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8</math> | ||
| − | ==Solution== | + | ==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 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. | ||
| + | ==Solution 2== | ||
| + | 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>). | ||
==Videos Explaining Solution== | ==Videos Explaining Solution== | ||
Revision as of 15:59, 28 July 2020
Problem 20
How many different real numbers 
satisfy the equation ![\[(x^{2}-5)^{2}=16?\]](http://latex.artofproblemsolving.com/d/1/8/d188c4697f3f7cf1889796a037bd60b6dae3b29e.png)
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 ).
Videos Explaining Solution
https://youtu.be/5BXh0JY4klM (Uses a difference of squares & factoring method, different from above solutions)
https://www.youtube.com/watch?v=44vrsk_CbF8&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=2 ~ MathEx
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. 
