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

(Solution 1)
(Solution 1)
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==Solution 1==
 
==Solution 1==
  
We know that to get 16, you can square 4 or -4. Thus, x squared - 5 can have 2 possibilities. x squared is either 9 or 1, leaving x with possiblities 3,-3, 1, and -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>.
  
 
==See Also==
 
==See Also==

Revision as of 19:41, 20 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 $\boxed{(D)}$.

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

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