AMC 10 · 2021 · #15
Grade 8 algebraProblem
The real number x satisfies the equation x+x1=5. What is the value of x11−7x7+x3?
Pick an answer.
AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A real number $x$ satisfies $x + \dfrac{1}{x} = \sqrt{5}$. Find the value of $x^{11} - 7 x^{7} + x^{3}$.
Givens: $x + \dfrac{1}{x} = \sqrt{5}$ (and $x \neq 0$); Target expression: $x^{11} - 7 x^{7} + x^{3}$; Answer choices: $-1,\; 0,\; 1,\; 2,\; \sqrt{5}$
Unknowns: The value of $x^{11} - 7 x^{7} + x^{3}$
Understand
Restated: A real number $x$ satisfies $x + \dfrac{1}{x} = \sqrt{5}$. Find the value of $x^{11} - 7 x^{7} + x^{3}$.
Givens: $x + \dfrac{1}{x} = \sqrt{5}$ (and $x \neq 0$); Target expression: $x^{11} - 7 x^{7} + x^{3}$; Answer choices: $-1,\; 0,\; 1,\; 2,\; \sqrt{5}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #9 Solve an Easier Related Problem, #13 Convert to Algebra, #3 Eliminate Possibilities
Tool #7 (Subproblems): the target $x^{11} - 7x^7 + x^3$ factors as $x^3(x^8 - 7x^4 + 1)$, splitting the work into 'find $x^4$ in simple form' and 'evaluate the bracket'. Tool #9 (Easier Problem): instead of working with the messy $\sqrt{5}$, square $x + 1/x = \sqrt{5}$ to get a polynomial relation, then multiply by $x^2$ to clear denominators. The result $x^4 = 3x^2 - 1$ is the easier version. Tool #13 (Algebra): substitute $u = x^4 = 3x^2 - 1$ into $u^2 - 7u + 1$ and watch all the $x^2$ terms cancel to $0$. Tool #3 (Eliminate) matches $0$ to choice (B).
Execute — Answer: B
7.EE.A.1 Step 1 - Factor the target expression to expose the structure.
- $x^{11} - 7x^7 + x^3 = x^3 (x^8 - 7 x^4 + 1)$.
- The bracket is a quadratic in $u = x^4$: $u^2 - 7u + 1$.
💡 Grade 7 expression manipulation: pull out the common factor $x^3$ to simplify.
8.EE.A.1 Step 2 - Square the given relation to kill $\sqrt{5}$.
- $\bigl(x + \tfrac{1}{x}\bigr)^2 = (\sqrt{5})^2 = 5$, so $x^2 + 2 + \tfrac{1}{x^2} = 5$ and therefore $x^2 + \tfrac{1}{x^2} = 3$.
💡 Grade 8 integer exponent rules: squaring removes the radical and gives a polynomial relation.
7.EE.A.2 Step 3 - Multiply by $x^2$ to clear the denominator.
- $x^4 + 1 = 3x^2$, i.e.
- $x^4 = 3x^2 - 1$.
- This is a clean expression for $u = x^4$ in terms of $x^2$.
💡 Grade 7 rewrite-to-reveal: a single variable $x^2$ now controls everything.
7.EE.A.1 Step 4 - Compute $u^2 = (3x^2 - 1)^2 = 9x^4 - 6x^2 + 1$.
- Replace the remaining $x^4$ using the same relation: $9x^4 = 9(3x^2 - 1) = 27x^2 - 9$.
- So $u^2 = 27x^2 - 9 - 6x^2 + 1 = 21x^2 - 8$.
💡 Grade 7 expand and substitute: keep reducing higher powers of $x$ to $x^2$.
7.EE.A.1 Step 5 - Plug $u^2 = 21x^2 - 8$ and $u = 3x^2 - 1$ into the bracket $u^2 - 7u + 1$.
- Get $(21x^2 - 8) - 7(3x^2 - 1) + 1 = 21x^2 - 8 - 21x^2 + 7 + 1 = 0$.
- The $x^2$ terms cancel exactly.
💡 Grade 7 combining like terms: the $x^2$ pieces add to $0$.
6.EE.A.4 Step 6 - Put it together.
- $x^{11} - 7x^7 + x^3 = x^3 \cdot (u^2 - 7u + 1) = x^3 \cdot 0 = 0$.
- (We don't need the explicit value of $x$; $x^3$ multiplies zero either way.) Match to choices: (B).
💡 Grade 6 equivalence: any number times $0$ is $0$, regardless of $x$.
7.EE.A.1 Factor the target expression to expose the structure. $x^{11} - 7x^7 + x^3 = x^3 8.EE.A.1 Square the given relation to kill $\sqrt{5}$. $\bigl(x + \tfrac{1}{x}\bigr)^2 = 7.EE.A.2 Multiply by $x^2$ to clear the denominator. $x^4 + 1 = 3x^2$, i.e. $x^4 = 3x^2 - 7.EE.A.1 Compute $u^2 = (3x^2 - 1)^2 = 9x^4 - 6x^2 + 1$. Replace the remaining $x^4$ usin 7.EE.A.1 Plug $u^2 = 21x^2 - 8$ and $u = 3x^2 - 1$ into the bracket $u^2 - 7u + 1$. Get $ 6.EE.A.4 Put it together. $x^{11} - 7x^7 + x^3 = x^3 \cdot (u^2 - 7u + 1) = x^3 \cdot 0 = Review
Reasonableness: Independent check: from $x + 1/x = \sqrt{5}$, the quadratic $x^2 - \sqrt{5}\,x + 1 = 0$ gives $x = \dfrac{\sqrt{5} \pm 1}{2}$ (a positive root of value $\dfrac{\sqrt{5} + 1}{2} \approx 1.618$, the golden ratio $\varphi$). Using $\varphi$: $\varphi^2 = \varphi + 1 \approx 2.618$, $\varphi^4 = (\varphi+1)^2 = \varphi^2 + 2\varphi + 1 \approx 6.854$. And $3\varphi^2 - 1 = 3(2.618) - 1 = 6.854$. So $x^4 = 3x^2 - 1$ holds. The bracket $x^8 - 7x^4 + 1 = (3x^2 - 1)^2 - 7(3x^2 - 1) + 1 = 0$ confirms the value $0$ on the nose. Answer (B) makes sense.
Alternative: Tool #5 (Pattern, via Fibonacci-style recursion) — building on $x + 1/x = \sqrt{5}$. Note $x^n + x^{-n}$ satisfies a recursion: let $a_n = x^n + x^{-n}$. Then $a_1 = \sqrt{5}$, $a_2 = 3$, and $a_{n+1} = \sqrt{5}\, a_n - a_{n-1}$. Compute $a_3 = 3\sqrt{5} - \sqrt{5} = 2\sqrt{5}$, $a_4 = 10 - 3 = 7$, etc. Factor target as $x^3(x^8 - 7x^4 + 1)$ and divide bracket by $x^4$: $x^4 - 7 + x^{-4} = a_4 - 7 = 7 - 7 = 0$. So the original expression is $x^3 \cdot x^4 \cdot 0 = 0$. Same answer (B).
CCSS standards used (min grade 8)
6.EE.A.4Identify when two expressions are equivalent (Concluding $x^3 \cdot 0 = 0$ regardless of $x$.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Factoring $x^{11} - 7x^7 + x^3 = x^3(x^8 - 7x^4 + 1)$ and expanding $(3x^2 - 1)^2$.)7.EE.A.2Rewrite an expression in different forms to shed light on the problem (Rewriting $x^2 + 1/x^2 = 3$ as $x^4 = 3x^2 - 1$.)8.EE.A.1Know and apply the properties of integer exponents (Squaring the given equation: $(x + 1/x)^2 = x^2 + 2 + 1/x^2$.)
⭐ This AMC 10 problem only needs Grade 8 exponent rules you already know! Factor the target: $x^{11} - 7x^7 + x^3 = x^3(x^8 - 7x^4 + 1)$. Squaring $x + 1/x = \sqrt{5}$ gives $x^2 + 1/x^2 = 3$, so $x^4 = 3x^2 - 1$. Plug into the bracket — every $x^2$ cancels and the bracket equals $0$. So the whole expression is $0$, answer (B).
⭐ This AMC 10 problem only needs Grade 8 exponent rules you already know! Factor the target: $x^{11} - 7x^7 + x^3 = x^3(x^8 - 7x^4 + 1)$. Squaring $x + 1/x = \sqrt{5}$ gives $x^2 + 1/x^2 = 3$, so $x^4 = 3x^2 - 1$. Plug into the bracket — every $x^2$ cancels and the bracket equals $0$. So the whole expression is $0$, answer (B).
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