Difference between revisions of "2021 AMC 10B Problems/Problem 15"
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==Solution 1== | ==Solution 1== | ||
− | We square <math>x+\frac{1}{x}=\sqrt5</math> to get <math>x^2+2+\frac{1}{x^2}=5</math>. We subtract 2 on both sides for <math>x^2+\frac{1}{x^2}=3</math> and square again, and see that <math>x^4+2+\frac{1}{x^4}=9</math> so <math>x^4+\frac{1}{x^4}=7</math>. We can | + | We square <math>x+\frac{1}{x}=\sqrt5</math> to get <math>x^2+2+\frac{1}{x^2}=5</math>. We subtract 2 on both sides for <math>x^2+\frac{1}{x^2}=3</math> and square again, and see that <math>x^4+2+\frac{1}{x^4}=9</math> so <math>x^4+\frac{1}{x^4}=7</math>. We can factor out <math>x^7</math> from our original expression of <math>x^{11}-7x^7+x^3</math> to get that it is equal to <math>x^7(x^4-7+\frac{1}{x^4})</math>. Therefore because <math>x^4+\frac{1}{x^4}</math> is 7, it is equal to <math>x^7(0)=\boxed{\textbf{(B) } 0}</math>. |
==Solution 2== | ==Solution 2== | ||
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&=(x^6-2x^5)+x^3 \ | &=(x^6-2x^5)+x^3 \ | ||
&=(-x^5+x^4+x^3) \ | &=(-x^5+x^4+x^3) \ | ||
− | &=-x^3(x^2-x-1) = \boxed{ | + | &=-x^3(x^2-x-1) = \boxed{\textbf{(B) } 0} |
\end{align*}</cmath> | \end{align*}</cmath> | ||
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==Solution 3== | ==Solution 3== | ||
− | We can immediately note that the exponents of <math>x^{11}-7x^7+x^3</math> are an arithmetic sequence, so they are symmetric around the middle term. So, <math>x^{11}-7x^7+x^3 = x^7(x^4-7+\frac{1}{x^4})</math>. We can see that since <math>x+\frac{1}{x} = \sqrt{5}</math>, <math>x^2+2+\frac{1}{x^2} = 5</math> and therefore <math>x^2+\frac{1}{x^2} = 3</math>. Continuing from here, we get <math>x^4+2+\frac{1}{x^4} = 9</math>, so <math>x^4-7+\frac{1}{x^4} = 0</math>. We don't even need to find what <math>x^ | + | We can immediately note that the exponents of <math>x^{11}-7x^7+x^3</math> are an arithmetic sequence, so they are symmetric around the middle term. So, <math>x^{11}-7x^7+x^3 = x^7(x^4-7+\frac{1}{x^4})</math>. We can see that since <math>x+\frac{1}{x} = \sqrt{5}</math>, <math>x^2+2+\frac{1}{x^2} = 5</math> and therefore <math>x^2+\frac{1}{x^2} = 3</math>. Continuing from here, we get <math>x^4+2+\frac{1}{x^4} = 9</math>, so <math>x^4-7+\frac{1}{x^4} = 0</math>. We don't even need to find what <math>x^7</math> is! This is since <math>x^7\cdot0</math> is evidently <math>\boxed{\textbf{(B) } 0}</math>, which is our answer. |
~sosiaops | ~sosiaops | ||
==Solution 4== | ==Solution 4== | ||
− | We begin by multiplying <math>x+\frac{1}{x} = \sqrt{5}</math> by <math>x</math>, resulting in <math>x^2+1 = \sqrt{5}x</math>. Now we see this equation: <math>x^{11}-7x^{7}+x^3</math>. The terms all have <math>x^3</math> in common, so we can factor that out, and what we're looking for becomes <math>x^3(x^8-7x^4+1)</math>. Looking back to our original equation, we have <math>x^2+1 = \sqrt{5}x</math>, which is equal to <math>x^2 = \sqrt{5}x-1</math>. Using this, we can evaluate <math>x^4</math> to be <math>5x^2-2\sqrt{5}x+1</math>, and we see that there is another <math>x^2</math>, so we put substitute it in again, resulting in <math>3\sqrt{5}x-4</math>. Using the same way, we find that <math>x^8</math> is <math> | + | We begin by multiplying <math>x+\frac{1}{x} = \sqrt{5}</math> by <math>x</math>, resulting in <math>x^2+1 = \sqrt{5}x</math>. Now we see this equation: <math>x^{11}-7x^{7}+x^3</math>. The terms all have <math>x^3</math> in common, so we can factor that out, and what we're looking for becomes <math>x^3(x^8-7x^4+1)</math>. Looking back to our original equation, we have <math>x^2+1 = \sqrt{5}x</math>, which is equal to <math>x^2 = \sqrt{5}x-1</math>. Using this, we can evaluate <math>x^4</math> to be <math>5x^2-2\sqrt{5}x+1</math>, and we see that there is another <math>x^2</math>, so we put substitute it in again, resulting in <math>3\sqrt{5}x-4</math>. Using the same way, we find that <math>x^8</math> is <math>21\sqrt{5}x-29</math>. We put this into <math>x^3(x^8-7x^4+1)</math>, resulting in <math>x^3(0)</math>, so the answer is <math>\boxed{(B)~0}</math>. |
~purplepenguin2 | ~purplepenguin2 | ||
− | == | + | ==Solution 5== |
− | https://youtu.be/ | + | The equation we are given is <math>x+\tfrac{1}{x}=\sqrt{5}...</math> Yuck. Fractions and radicals! We multiply both sides by <math>x,</math> square, and re-arrange to get <cmath>x^2+1=\sqrt{5}x \implies x^4+2x^2+1=5x^2 \implies x^4-3x^2+1=0.</cmath> Now, let us consider the expression we wish to acquire. Factoring out <math>x^3,</math> we have <cmath>x^3\left(x^8-7x^4+1\right) = x^3\left(x^8+2x^4+1-9x^4\right).</cmath> Then, we notice that <math>x^8+2x^4+1=\left(x^4+1\right)^2.</math> Furthermore, <cmath>x^4+1=3x^2 \implies \left(x^4+1\right)^2=x^8+2x^4+1=9x^4.</cmath> Thus, our answer is <cmath>x^3\left(9x^4-9x^4\right) = x^3 \cdot 0 = \boxed{\textbf{(B)}} ~ 0.</cmath> |
+ | ~peace09 | ||
+ | ==Solution 6(Non-rigorous for little time)== | ||
+ | Multiplying by x and solving, we get that <math>x = \frac{\sqrt{5} \pm 1}{2}.</math> Note that whether or not we take <math>x = \frac{\sqrt{5} + 1}{2}</math> or we take <math>\frac{\sqrt{5} - 1}{2},</math> our answer has to be the same. Thus, we take <math>x = \frac{\sqrt{5} - 1}{2} \approx 0.62</math>. Since this number is small, taking it to high powers like <math>11</math>, <math>7</math>, and <math>3</math> will make the number very close to <math>0</math>, so the answer is <math>\boxed{(B)~0}.</math> | ||
+ | ~AtharvNaphade | ||
+ | |||
+ | ==Solution 7== | ||
+ | We know that <math>x+\frac{1}{x}=\sqrt{5}</math>. Multiply both sides by <math>x</math> to get <math>x^2+1=x\sqrt{5}</math> | ||
+ | Squaring both sides: <cmath>x^4+2x^2+1=5x^2</cmath> | ||
+ | Subtract <math>2x^2</math> from both sides: <cmath>x^4+1=3x^2</cmath> | ||
+ | Squaring both sides: <cmath>x^8+2x^4+1=9x^4</cmath> | ||
+ | Subtract <math>9x^4</math> from both sides: <cmath>x^8-7x^4+1=0</cmath> | ||
+ | Multiply <math>x^3</math> on both sides: <cmath>x^{11}-7x^7+x^3=\fbox{(B) 0}</cmath> | ||
+ | ~sid2012 [https://artofproblemsolving.com/wiki/index.php/User:Sid2012] | ||
+ | |||
+ | ==Video Solution (🚀 Super Fast. Under 2 min! 🚀)== | ||
+ | https://youtu.be/CJbtpNhMvIM | ||
+ | |||
+ | <i> ~Education, the Study of Everything </i> | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/M4Ffhp9NLKY?t=81 | ||
~ pi_is_3.14 | ~ pi_is_3.14 |
Latest revision as of 13:06, 29 September 2024
Contents
[hide]- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6(Non-rigorous for little time)
- 8 Solution 7
- 9 Video Solution (🚀 Super Fast. Under 2 min! 🚀)
- 10 Video Solution by OmegaLearn
- 11 Video Solution by Interstigation (Simple Silly Bashing)
- 12 Video Solution by TheBeautyofMath
- 13 See Also
Problem
The real number satisfies the equation . What is the value of
Solution 1
We square to get . We subtract 2 on both sides for and square again, and see that so . We can factor out from our original expression of to get that it is equal to . Therefore because is 7, it is equal to .
Solution 2
Multiplying both sides by and using the quadratic formula, we get . We can assume that it is , and notice that this is also a solution the equation , i.e. we have . Repeatedly using this on the given (you can also just note Fibonacci numbers),
~Lcz
Solution 3
We can immediately note that the exponents of are an arithmetic sequence, so they are symmetric around the middle term. So, . We can see that since , and therefore . Continuing from here, we get , so . We don't even need to find what is! This is since is evidently , which is our answer.
~sosiaops
Solution 4
We begin by multiplying by , resulting in . Now we see this equation: . The terms all have in common, so we can factor that out, and what we're looking for becomes . Looking back to our original equation, we have , which is equal to . Using this, we can evaluate to be , and we see that there is another , so we put substitute it in again, resulting in . Using the same way, we find that is . We put this into , resulting in , so the answer is .
~purplepenguin2
Solution 5
The equation we are given is Yuck. Fractions and radicals! We multiply both sides by square, and re-arrange to get Now, let us consider the expression we wish to acquire. Factoring out we have Then, we notice that Furthermore, Thus, our answer is ~peace09
Solution 6(Non-rigorous for little time)
Multiplying by x and solving, we get that Note that whether or not we take or we take our answer has to be the same. Thus, we take . Since this number is small, taking it to high powers like , , and will make the number very close to , so the answer is ~AtharvNaphade
Solution 7
We know that . Multiply both sides by to get Squaring both sides: Subtract from both sides: Squaring both sides: Subtract from both sides: Multiply on both sides: ~sid2012 [1]
Video Solution (🚀 Super Fast. Under 2 min! 🚀)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/M4Ffhp9NLKY?t=81
~ pi_is_3.14
Video Solution by Interstigation (Simple Silly Bashing)
~ Interstigation
Video Solution by TheBeautyofMath
Not the most efficient method, but gets the job done.
https://youtu.be/L1iW94Ue3eI?t=1468
~IceMatrix
See Also
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.