Difference between revisions of "2019 AMC 8 Problems/Problem 20"
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− | ==Problem | + | ==Problem== |
How many different real numbers <math>x</math> satisfy the equation <cmath>(x^{2}-5)^{2}=16?</cmath> | How many different real numbers <math>x</math> satisfy the equation <cmath>(x^{2}-5)^{2}=16?</cmath> | ||
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==Solution 1== | ==Solution 1== | ||
− | We have that <math>(x^2-5)^2 = 16</math> if and only if <math>x^2-5 = \pm 4</math>. If <math>x^2-5 = 4</math>, then <math>x^2 = 9 \implies x = \pm 3</math>, giving 2 solutions. If <math>x^2-5 = -4</math>, then <math>x^2 = 1 \implies x = \pm 1</math>, giving 2 more solutions. All four of these solutions work, so the answer is <math>\boxed{\textbf{(D)} 4}</math>. Further, the equation is a quartic in <math>x</math>, so by the [https://artofproblemsolving.com/wiki/index.php/Fundamental_Theorem_of_Algebra Fundamental Theorem of Algebra], there can be at most four real solutions. | + | We have that <math>(x^2-5)^2 = 16</math> if and only if <math>x^2-5 = \pm 4</math>. If <math>x^2-5 = 4</math>, then <math>x^2 = 9 \implies x = \pm 3</math>, giving 2 solutions. If <math>x^2-5 = -4</math>, then <math>x^2 = 1 \implies x = \pm 1</math>, giving 2 more solutions. All four of these solutions work, so the answer is <math>\boxed{\textbf{(D) }4}</math>. Further, the equation is a [[quartic Equation|quartic]] in <math>x</math>, so by the [https://artofproblemsolving.com/wiki/index.php/Fundamental_Theorem_of_Algebra Fundamental Theorem of Algebra], there can be at most four real solutions. |
==Solution 2== | ==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>. | + | 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 | + | ~UnstoppableGoddess |
+ | |||
+ | ==Solution 3== | ||
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+ | Subtract 16 from both sides and factor using difference of squares: | ||
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+ | <cmath>(x^2 - 5)^2 = 16 </cmath> | ||
+ | <cmath>(x^2 - 5)^2 - 16 =0 </cmath> | ||
+ | <cmath>(x^2 - 5)^2 - 4^2 = 0 </cmath> | ||
+ | <cmath>[(x^2 - 5)-4][(x^2 - 5) + 4] = 0</cmath> | ||
+ | <cmath>(x^2 - 9)(x^2 - 1) =0 </cmath> | ||
+ | <cmath>(x+3)(x-3)(x+1)(x-1) = 0 </cmath> | ||
+ | |||
+ | |||
+ | Quite obviously, this equation has <math>\boxed{\textbf{(D) }4}</math> solutions. | ||
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+ | |||
+ | ~TaeKim | ||
+ | |||
+ | ==Video Solution by Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/IgpayYB48C4?si=EHbnc8zZoQ15Gfv6&t=6050 | ||
+ | |||
+ | ~Math-X | ||
==Solution 3== | ==Solution 3== | ||
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https://youtu.be/5BXh0JY4klM (Uses a difference of squares & factoring method, different from above solutions) | https://youtu.be/5BXh0JY4klM (Uses a difference of squares & factoring method, different from above solutions) | ||
− | ==Solution 5 ( | + | ==Solution 5 (video of Solution 1)== |
https://www.youtube.com/watch?v=44vrsk_CbF8&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=2 ~ MathEx | https://www.youtube.com/watch?v=44vrsk_CbF8&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=2 ~ MathEx | ||
− | ==Solution | + | == Video Solution by Pi Academy == |
− | Video Solution - https://youtu.be/ | + | |
+ | https://youtu.be/Ds8Nzjj6pXs?si=QAwrO_bZHrTj6cba | ||
+ | |||
+ | ~ smartschoolboy9 | ||
+ | |||
+ | == Video Solution 2 (Gateway to Harder Questions)== | ||
+ | |||
+ | https://www.youtube.com/watch?v=J-E4SGEi3QE&t=2s | ||
+ | |||
+ | https://youtu.be/V3HxkJhSn08 -Happytwin | ||
+ | |||
+ | Solution detailing how to solve the problem: https://youtu.be/x4cF3o3Fzj8 | ||
+ | |||
+ | https://youtu.be/dh9uf5_ZB5Q ~ Education, the Study of Everything | ||
+ | |||
+ | https://youtu.be/Xm4ZGND9WoY ~ Hayabusa1 | ||
− | + | https://www.youtube.com/watch?v=TpsuRedYOiM&t=250s ~ SpreadTheMathLove | |
− | https:// | ||
− | |||
==See Also== | ==See Also== |
Latest revision as of 09:15, 9 November 2024
Contents
Problem
How many different real numbers satisfy the equation
Solution 1
We have that if and only if . If , then , giving 2 solutions. If , then , giving 2 more solutions. All four of these solutions work, so the answer is . Further, the equation is a quartic in , so by the Fundamental Theorem of Algebra, there can be at most four real solutions.
Solution 2
We can expand to get , so now our equation is . Subtracting from both sides gives us . Now, we can factor the left hand side to get . If and/or equals , then the whole left side will equal . Since the solutions can be both positive and negative, we have solutions: (we can find these solutions by setting and equal to and solving for ). So, the answer is .
~UnstoppableGoddess
Solution 3
Subtract 16 from both sides and factor using difference of squares:
Quite obviously, this equation has solutions.
~TaeKim
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/IgpayYB48C4?si=EHbnc8zZoQ15Gfv6&t=6050
~Math-X
Solution 3
Associated Video - https://www.youtube.com/watch?v=Q5yfodutpsw
Solution 4
https://youtu.be/5BXh0JY4klM (Uses a difference of squares & factoring method, different from above solutions)
Solution 5 (video of Solution 1)
https://www.youtube.com/watch?v=44vrsk_CbF8&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=2 ~ MathEx
Video Solution by Pi Academy
https://youtu.be/Ds8Nzjj6pXs?si=QAwrO_bZHrTj6cba
~ smartschoolboy9
Video Solution 2 (Gateway to Harder Questions)
https://www.youtube.com/watch?v=J-E4SGEi3QE&t=2s
https://youtu.be/V3HxkJhSn08 -Happytwin
Solution detailing how to solve the problem: https://youtu.be/x4cF3o3Fzj8
https://youtu.be/dh9uf5_ZB5Q ~ Education, the Study of Everything
https://youtu.be/Xm4ZGND9WoY ~ Hayabusa1
https://www.youtube.com/watch?v=TpsuRedYOiM&t=250s ~ SpreadTheMathLove
See Also
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.