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

(Solution 1)
(Solution 1)
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==Solution 1==
 
==Solution 1==
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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.
  
 
==See Also==
 
==See Also==

Revision as of 20:21, 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 squared - 5 can have 2 possibilities. x squared is either 9 or 1, leaving x with possiblities 3,-3, 1, and -1.

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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