AMC 8 Daily Practice Round 9

Complete problem set with solutions and individual problem pages

Problem 8 Easy

Given that {x^{2}-6x+1=0}, and x \ne 0. Find the value of {x^{4}+\dfrac{1}{x^{4}}}.

  • A.

    1131

  • B.

    1154

  • C.

    1156

  • D.

    1197

  • E.

    1290

Answer:B

{{x}^{2}}-6x+1=0

x-6+\frac{1}{x}=0（∵x\ne 0）

x+\frac{1}{x}=6

{{\left( x+\frac{1}{x} \right)}^{2}}={{6}^{2}}

{{x}^{2}}+\frac{1}{{{x}^{2}}}=36-2

~~~~~~~~~~~~~~~=34

{{\left( x+\frac{1}{x} \right)}^{4}}={{6}^{4}}

{{x}^{4}}+4{{x}^{2}}+6+\frac{4}{{{x}^{2}}}+\frac{1}{{{x}^{4}}}=1296

\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)+4\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)=1290

{{x}^{4}}+\frac{1}{{{x}^{4}}}=1290-136

~~~~~~~~~~~~~~~=1154．

The answer is \text{B}.

Related Problems
Problem 6 AMC 8 Daily Practice Round 9
Problem 7 AMC 8 Daily Practice Round 9
Problem 9 AMC 8 Daily Practice Round 9
Problem 10 AMC 8 Daily Practice Round 9
Think Academy Powered by ChatGPT