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

(Solution 2 is incorrect as the fact that it is a quartic does not imply all 4 solutions are real.)
(→‎Solution 1)
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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>.
 
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>.
 +
==Solution 2==
 +
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>
  
 
==See Also==
 
==See Also==

Revision as of 14:09, 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 1

We know that to get $16$, you can square $4$ or $-4$. Thus, $x^2 - 5$ can have $2$ possibilities. $x^2$ is either $9$ or $1$, leaving $x$ with possiblities $3,-3, 1,$ and $-1,$ so $\boxed{(D)}$.

Solution 2

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 $\boxed{(D), 4}$

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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