Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 AMC 8 Problems/Problem 20"

(→‎Solution 5 (video of Solution 1))
(→‎Video Solution 1)
Line 43: Line 43:
 
https://www.youtube.com/watch?v=44vrsk_CbF8&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=2 ~ MathEx
 
https://www.youtube.com/watch?v=44vrsk_CbF8&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=2 ~ MathEx
  
−
==Video Solution 1==
+
==Video Solution 1 (Gateway to Harder Questions)==
  
 
https://www.youtube.com/watch?v=J-E4SGEi3QE&t=2s
 
https://www.youtube.com/watch?v=J-E4SGEi3QE&t=2s

Revision as of 20:48, 26 December 2023

Problem 20

How many different real numbers $x$ satisfy the equation \[(x^{2}-5)^{2}=16?\]

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8$

Solution 1

We have that $(x^2-5)^2 = 16$ if and only if $x^2-5 = \pm 4$. If $x^2-5 = 4$, then $x^2 = 9 \implies x = \pm 3$, giving 2 solutions. If $x^2-5 = -4$, then $x^2 = 1 \implies x = \pm 1$, giving 2 more solutions. All four of these solutions work, so the answer is $\boxed{\textbf{(D) }4}$. Further, the equation is a quartic in $x$, so by the Fundamental Theorem of Algebra, there can be at most four real solutions.

Solution 2

We can expand $(x^2-5)^2$ to get $x^4-10x^2+25$, so now our equation is $x^4-10x^2+25=16$. Subtracting $16$ from both sides gives us $x^4-10x^2+9=0$. Now, we can factor the left hand side to get $(x^2-9)(x^2-1)=0$. If $x^2-9$ and/or $x^2-1$ equals $0$, then the whole left side will equal $0$. Since the solutions can be both positive and negative, we have $4$ solutions: $-3,3,-1,1$ (we can find these solutions by setting $x^2-9$ and $x^2-1$ equal to $0$ and solving for $x$). So, the answer is $\boxed{\textbf{(D) }4}$.

~UnstoppableGoddess

Solution 3

Subtract 16 from both sides and factor using difference of squares:


\[(x^2 - 5)^2 = 16\] \[(x^2 - 5)^2 - 16 =0\] \[(x^2 - 5)^2 - 4^2 = 0\] \[[(x^2 - 5)-4][(x^2 - 5) + 4] = 0\] \[(x^2 - 9)(x^2 - 1) =0\] \[(x+3)(x-3)(x+1)(x-1) = 0\]


Quite obviously, this equation has $\boxed{\textbf{(D) }4}$ solutions.


~TaeKim

Solution 3

Associated Video - https://www.youtube.com/watch?v=Q5yfodutpsw

https://youtu.be/0AY1klX3gBo

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 1 (Gateway to Harder Questions)

https://www.youtube.com/watch?v=J-E4SGEi3QE&t=2s

Video Solution

https://youtu.be/V3HxkJhSn08

-Happytwin

Video Solution

Solution detailing how to solve the problem: https://youtu.be/x4cF3o3Fzj8

Video Solution

https://youtu.be/dh9uf5_ZB5Q

~Education, the Study of Everything

Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_20&oldid=208466"