Difference between revisions of "2021 AMC 10B Problems/Problem 15"

Latest revision as of 17:40, 1 November 2025

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 Solution 8 (very intuitive & efficient)
10 Video Solution (Super Fast. 2 min and 9 seconds)
11 Video Solution (Easiest and Fastest BASIC UNDERSTANDING ONLY Required)
12 Video Solution by OmegaLearn
13 Video Solution by Interstigation (Simple Silly Bashing)
14 Video Solution by TheBeautyofMath
15 See Also

Problem

The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}$ . What is the value of $x^{11}-7x^{7}+x^3?$

$\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}$

Solution 1

We square $x+\frac{1}{x}=\sqrt5$ to get $x^2+2+\frac{1}{x^2}=5$ . We subtract 2 on both sides for $x^2+\frac{1}{x^2}=3$ and square again, and see that $x^4+2+\frac{1}{x^4}=9$ so $x^4+\frac{1}{x^4}=7$ . We can factor out $x^7$ from our original expression of $x^{11}-7x^7+x^3$ to get that it is equal to $x^7(x^4-7+\frac{1}{x^4})$ . Therefore because $x^4+\frac{1}{x^4}$ is 7, it is equal to $x^7(0)=\boxed{\textbf{(B) } 0}$ .

Solution 2

Multiplying both sides by $x$ and using the quadratic formula, we get $\frac{\sqrt{5} \pm 1}{2}$ . We can assume that it is $\frac{\sqrt{5}+1}{2}$ , and notice that this is the golden mean $\varphi$ , which is well-known to be a solution the equation $x^2-x-1=0$ , i.e. we have $x^2=x+1$ . Repeatedly using this on the given (you can also just note Fibonacci numbers), $\begin{align*} (x^{11})-7x^7+x^3 &= (x^{10}+x^9)-7x^7+x^3 \\ &=(2x^9+x^8)-7x^7+x^3 \\ &=(3x^8+2x^7)-7x^7+x^3 \\ &=(3x^8-5x^7)+x^3 \\ &=(-2x^7+3x^6)+x^3 \\ &=(x^6-2x^5)+x^3 \\ &=(-x^5+x^4+x^3) \\ &=-x^3(x^2-x-1) = \boxed{\textbf{(B) } 0} \end{align*}$

~Lcz

Solution 3

We can immediately note that the exponents of $x^{11}-7x^7+x^3$ are an arithmetic sequence, so they are symmetric around the middle term. So, $x^{11}-7x^7+x^3 = x^7(x^4-7+\frac{1}{x^4})$ . We can see that since $x+\frac{1}{x} = \sqrt{5}$ , $x^2+2+\frac{1}{x^2} = 5$ and therefore $x^2+\frac{1}{x^2} = 3$ . Continuing from here, we get $x^4+2+\frac{1}{x^4} = 9$ , so $x^4-7+\frac{1}{x^4} = 0$ . We don't even need to find what $x^7$ is! This is since $x^7\cdot0$ is evidently $\boxed{\textbf{(B) } 0}$ , which is our answer.

~sosiaops

Solution 4

We begin by multiplying $x+\frac{1}{x} = \sqrt{5}$ by $x$ , resulting in $x^2+1 = \sqrt{5}x$ . Now we see this equation: $x^{11}-7x^{7}+x^3$ . The terms all have $x^3$ in common, so we can factor that out, and what we're looking for becomes $x^3(x^8-7x^4+1)$ . Looking back to our original equation, we have $x^2+1 = \sqrt{5}x$ , which is equal to $x^2 = \sqrt{5}x-1$ . Using this, we can evaluate $x^4$ to be $5x^2-2\sqrt{5}x+1$ , and we see that there is another $x^2$ , so we put substitute it in again, resulting in $3\sqrt{5}x-4$ . Using the same way, we find that $x^8$ is $21\sqrt{5}x-29$ . We put this into $x^3(x^8-7x^4+1)$ , resulting in $x^3(0)$ , so the answer is $\boxed{(B)~0}$ .

~purplepenguin2

Solution 5

The equation we are given is $x+\tfrac{1}{x}=\sqrt{5}...$ Yuck. Fractions and radicals! We multiply both sides by $x,$ square, and re-arrange to get $x^2+1=\sqrt{5}x \implies x^4+2x^2+1=5x^2 \implies x^4-3x^2+1=0.$ Now, let us consider the expression we wish to acquire. Factoring out $x^3,$ we have $x^3\left(x^8-7x^4+1\right) = x^3\left(x^8+2x^4+1-9x^4\right).$ Then, we notice that $x^8+2x^4+1=\left(x^4+1\right)^2.$ Furthermore, $x^4+1=3x^2 \implies \left(x^4+1\right)^2=x^8+2x^4+1=9x^4.$ Thus, our answer is $x^3\left(9x^4-9x^4\right) = x^3 \cdot 0 = \boxed{\textbf{(B)}} ~ 0.$ ~peace09

Solution 6(Non-rigorous for little time)

Multiplying by x and solving, we get that $x = \frac{\sqrt{5} \pm 1}{2}.$ Note that whether or not we take $x = \frac{\sqrt{5} + 1}{2}$ or we take $\frac{\sqrt{5} - 1}{2},$ our answer has to be the same. Thus, we take $x = \frac{\sqrt{5} - 1}{2} \approx 0.62$ . Since this number is small, taking it to high powers like $11$ , $7$ , and $3$ will make the number very close to $0$ , so the answer is $\boxed{(B)~0}.$ ~AtharvNaphade

Solution 7

We know that $x+\frac{1}{x}=\sqrt{5}$ . Multiply both sides by $x$ to get $x^2+1=x\sqrt{5}$ Squaring both sides: $x^4+2x^2+1=5x^2$ Subtract $2x^2$ from both sides: $x^4+1=3x^2$ Squaring both sides: $x^8+2x^4+1=9x^4$ Subtract $9x^4$ from both sides: $x^8-7x^4+1=0$ Multiply $x^3$ on both sides: $x^{11}-7x^7+x^3=\fbox{(B) 0}$ ~sid2012 [1]

Solution 8 (very intuitive & efficient)

Squaring $x+\frac{1}{x}=\sqrt{5}$ yields $x^2+2+\frac{1}{x^2}=5$ . We subtract $2$ from both sides, yielding $x^2+\frac{1}{x^2}=3$ , and, squaring again, we end up with $x^4+2+\frac{1}{x^4}=9$ . Subtracting $2$ from both sides again, we end up with $x^4+\frac{1}{x^4}=7$ . Observe that $7$ is the coefficient of the $x^7$ term in our second equation, $x^{11}-7x^{7}+x^3$ . We can now substitute $x^4+\frac{1}{x^4}$ into the second equation to result in $x^{11}-\left(x^4+\frac{1}{x^4}\right)x^{7}+x^3$ . Multiplying the middle term out yields $x^{11}-x^{11}-x^{3}+x^{3}$ , and all of the terms cancel out. Therefore, the answer is $\boxed{(B)~0}.$ ~rxm0203

Video Solution (Super Fast. 2 min and 9 seconds)

https://youtu.be/CJbtpNhMvIM

~Education, the Study of Everything

Video Solution (Easiest and Fastest BASIC UNDERSTANDING ONLY Required)

https://www.youtube.com/watch?v=tSTn4K-ZB20

Video Solution by OmegaLearn

https://youtu.be/M4Ffhp9NLKY?t=81

~ pi_is_3.14

Video Solution by Interstigation (Simple Silly Bashing)

https://youtu.be/Hdk2SDOcw7c

~ Interstigation

Video Solution by TheBeautyofMath

Not the most efficient method, but gets the job done.

https://youtu.be/L1iW94Ue3eI?t=1468

~IceMatrix

@@ Line 51: / Line 51: @@
 ~sid2012 [https://artofproblemsolving.com/wiki/index.php/User:Sid2012]
-==Solution 8 (very efficient)==
+==Solution 8 (very intuitive & efficient)==
-Squaring <math>x+\frac{1}{x}=\sqrt{5}</math> yields <math>x^2+2+\frac{1}{x^2}=5</math>. We subtract <math>2</math> from both sides, yielding <math>x^2+\frac{1}{x^2}=3</math>, and, squaring again, we end up with <math>x^4+2+\frac{1}{x^4}=9</math>. Subtracting <math>2</math> from both sides again, we end up with <math>x^4+\frac{1}{x^4}=7</math>. Observe that <math>7</math> is the coefficient of the <math>x^7</math> term in our second equation, <math>x^{11}-7x^{7}+x^3</math>. We can now substitute <math>x^4+\frac{1}{x^4}</math> into the second equation to result in <math>x^{11}-\left(x^4+\frac{1}{x^4}\right)x^{7}+x^3</math>. Multiplying the middle term out yields <math>x^{11}-x^{11}-x^{3}+x^{3}</math>, and all of the terms cancel out. Therefore, the answer is <math>\boxed{(B)~0}.</math>
+Squaring <math>x+\frac{1}{x}=\sqrt{5}</math> yields <math>x^2+2+\frac{1}{x^2}=5</math>. We subtract <math>2</math> from both sides, yielding <math>x^2+\frac{1}{x^2}=3</math>, and, squaring again, we end up with <math>x^4+2+\frac{1}{x^4}=9</math>. Subtracting <math>2</math> from both sides again, we end up with <math>x^4+\frac{1}{x^4}=7</math>. Observe that <math>7</math> is the coefficient of the <math>x^7</math> term in our second equation, <math>x^{11}-7x^{7}+x^3</math>. We can now substitute <math>x^4+\frac{1}{x^4}</math> into the second equation to result in <math>x^{11}-\left(x^4+\frac{1}{x^4}\right)x^{7}+x^3</math>. Multiplying the middle term out yields <math>x^{11}-x^{11}-x^{3}+x^{3}</math>, and all of the terms cancel out. Therefore, the answer is <math>\boxed{(B)~0}.</math> ~rxm0203
 ==Video Solution (Super Fast. 2 min and 9 seconds)==

2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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