2021 AMC 10B Problems/Problem 15: Difference between revisions

Latest revision as of 17:40, 1 November 2025

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 7: / Line 7: @@
 ==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 divide our original expression of <math>x^{11}-7x^7+x^3</math> by <math>x^7</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{(B) 0}</math>.
+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==
-Multiplying both sides by <math>x</math> and using the quadratic formula, we get <math>\frac{\sqrt{5} \pm 1}{2}</math>. We can assume that it is <math>\frac{\sqrt{5}+1}{2}</math>, and notice that this is also a solution the equation <math>x^2-x-1=0</math>, i.e. we have <math>x^2=x+1</math>. Repeatedly using this on the given (you can also just note Fibonacci numbers),
+Multiplying both sides by <math>x</math> and using the quadratic formula, we get <math>\frac{\sqrt{5} \pm 1}{2}</math>. We can assume that it is <math>\frac{\sqrt{5}+1}{2}</math>, and notice that this is the golden mean <math>\varphi</math>, which is well-known to be a solution the equation <math>x^2-x-1=0</math>, i.e. we have <math>x^2=x+1</math>. Repeatedly using this on the given (you can also just note Fibonacci numbers),
 <cmath> \begin{align*}
 (x^{11})-7x^7+x^3 &= (x^{10}+x^9)-7x^7+x^3 \\
@@ Line 20: / Line 20: @@
 &=(x^6-2x^5)+x^3 \\
 &=(-x^5+x^4+x^3) \\
-&=-x^3(x^2-x-1) = \boxed{(\textbf{B}) 0}
+&=-x^3(x^2-x-1) = \boxed{\textbf{(B) } 0}
 \end{align*}</cmath>
 ~Lcz
+==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^7</math> is! This is since <math>x^7\cdot0</math> is evidently <math>\boxed{\textbf{(B) } 0}</math>, which is our answer.
+~sosiaops
+==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>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
+==Solution 5==
+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]
+==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> ~rxm0203
+==Video Solution (Super Fast. 2 min and 9 seconds)==
+https://youtu.be/CJbtpNhMvIM
+<i> ~Education, the Study of Everything </i>
+==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
+==See Also==
 {{AMC10 box|year=2021|ab=B|num-b=14|num-a=16}}
+{{MAA Notice}}

2021 AMC 10B (Problems • Answer Key • Resources)
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