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

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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 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.

Solution 2

We can expand $(x^2-5)^2$ to get $x^4-10x^2+25$ , so now our equation is $x^4-10x^2+25=16$ . Subtracting $16$ from both sides gives us $x^4-10x^2+9=0$ . Now, we can factor the left hand side to get $(x^2-9)(x^2-1)=0$ . If $x^2-9$ and/or $x^2-1$ equals $0$ , then the whole left side will equal $0$ . Since the solutions can be both positive and negative, we have $4$ solutions: $-3,3,-1,1$ (we can find these solutions by setting $x^2-9$ and $x^2-1$ equal to $0$ and solving for $x$ ). So the answer is $\boxed{\textbf{(D)} 4}$ . ~UnstoppableGoddess

Videos Explaining Solution

https://youtu.be/0AY1klX3gBo

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

https://www.youtube.com/watch?v=44vrsk_CbF8&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=2 ~ MathEx

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 8: / Line 8: @@
 ==Solution 2==
-We can expand <math>(x^2-5)^2</math> to get <math>x^4-10x^2+25</math>, so now our equation is <math>x^4-10x^2+25=16</math>. Subtracting <math>16</math> from both sides gives us <math>x^4-10x^2+9=0</math>. Now, we can factor the left hand side to get <math>(x^2-9)(x^2-1)=0</math>. If <math>x^2-9</math> and/or <math>x^2-1</math> equals <math>0</math>, then the whole left side will equal <math>0</math>. Since the solutions can be both positive and negative, we have <math>4</math> solutions: <math>-3,3,-1,1</math> (we can find these solutions by setting <math>x^2-9</math> and <math>x^2-1</math> equal to <math>0</math> and solving for <math>x</math>). So the answer is <math>\boxed{\textbf{(D)} 4}</math>. ~UnstoppableGoddess
+We can expand <math>(x^2-5)^2</math> to get <math>x^4-10x^2+25</math>, so now our equation is <math>x^4-10x^2+25=16</math>. Subtracting <math>16</math> from both sides gives us <math>x^4-10x^2+9=0</math>. Now, we can factor the left hand side to get <math>(x^2-9)(x^2-1)=0</math>. If <math>x^2-9</math> and/or <math>x^2-1</math> equals <math>0</math>, then the whole left side will equal <math>0</math>. Since the solutions can be both positive and negative, we have <math>4</math> solutions: <math>-3,3,-1,1</math> (we can find these solutions by setting <math>x^2-9</math> and <math>x^2-1</math> equal to <math>0</math> and solving for <math>x</math>). So the answer is <math>\boxed{\textbf{(D)} 4}</math>.
+~UnstoppableGoddess
 ==Videos Explaining Solution==

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