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

(→‎Solution 2)
(condense solutions 1 and 2 to a more complete solution. Simply stating that the equation is quartic is not sufficient to show there are 4 real solutions.)
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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>
  
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==Solution 1==
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==Solution==
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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{(B)} 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.
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We know that to get <math>16</math>, you can square <math>4</math> or <math>-4</math>. Thus, <math>x^2 - 5</math> can have <math>2</math> possibilities. <math>x^2</math> is either <math>9</math> or <math>1</math>, leaving <math>x</math> with possiblities <math>3,-3, 1,</math> and <math>-1,</math> so <math>\boxed{(D)}</math>.
 
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==Solution 2==
 
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The equation is a quartic, so there will be 4 solutions. By skimming the problem, none of the answers are "extranerous", so the answer is <math>\boxed{(D) 4}</math>~heeeeeeeeheeeee
 
  
 
==See Also==
 
==See Also==

Revision as of 17:08, 22 November 2019

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

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 $\boxed{\textbf{(B)} 4}$. Further, the equation is a quartic in $x$, so by the fundamental theorem of algebra, there can be at most four real solutions.

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

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