Canonical URI: https://w3id.org/kmath/textbook_solution/visang_solution_p158_workbook_p137_10
| rdf:type | math:TextbookSolution |
|---|---|
| rdfs:label | 비상 p137 수학 익힘책 II-10 풀이 |
| rdfs:comment | 비상 공통수학1 정답 및 해설 p158의 수학 익힘책 p137 II-10 풀이. |
| math:answerText | \(4\) |
| math:explanationText | \[ (x-1)(x-2)(x+3)(x+4)=50 \] 에서 \[ (x^2+2x-3)(x^2+2x-8)-50=0 \] \(x^2+2x=X\)로 놓으면 \[ (X-3)(X-8)-50=0,\quad (X+2)(X-13)=0 \] 따라서 \[ (x^2+2x+2)(x^2+2x-13)=0 \] 즉, \(x=-1\pm i\) 또는 \(x=-1\pm\sqrt{14}\). 따라서 두 허근 \(\alpha,\ \beta\)에 대하여 \(\alpha+\beta=-2,\ \alpha\beta=2\)이므로 \[ \alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=4 \] |
| math:mappingConfidence | 1.0 |
| math:pageStart | 158 |
| math:problem | textbook_problem:visang_workbook_p137_10 |
| math:reviewStatus | reviewed |
| math:solutionKind | worked_solution |
| math:usesSolutionPattern | solution_pattern:roots_coefficients_symmetric_expression |
| math:usesSolutionPattern | solution_pattern:substitution_factorization |