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

(Solution 1)
(Solution 1)
Line 5: Line 5:
  
 
==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{(e)} 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==
 
==Solution 2==

Revision as of 00:55, 2 November 2020

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

https://youtu.be/0AY1klX3gBo

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

Solution 6

Video Solution - https://youtu.be/Lw8fSbX_8FU

Video Solution

https://youtu.be/V3HxkJhSn08

-Happytwin

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS