2019 AMC 8 Problems/Problem 20

Revision as of 14:09, 22 November 2019 by Heeeeeeeheeeee (talk | contribs) (Solution 1)

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AJHSME/AMC 8 Problems and Solutions

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