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

Revision as of 16:42, 27 May 2020

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 $\boxed{\textbf{(D)} 4}$ . Further, the equation is a quartic in $x$ , so by the Fundamental Theorem of Algebra, there can be at most four real solutions.

Videos Explaining Solution

https://youtu.be/0AY1klX3gBo https://youtu.be/5BXh0JY4klM (Uses a difference of squares & factoring method, different from above solutions)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 10: / Line 10: @@
 ==Videos Explaining Solution==
 https://youtu.be/0AY1klX3gBo
+https://youtu.be/5BXh0JY4klM (Uses a difference of squares & factoring method, different from above solutions)
 ==See Also==

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

Revision as of 16:42, 27 May 2020

Contents

Problem 20

Solution

Videos Explaining Solution

See Also