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Construct a two-column proof to prove the following by using factoring: Given: a³ - 11a²-9a = -99 Prove: a = 11, +3

Construct a two-column proof to prove the following by using factoring: Given: a³ - 11a²-9a = -99 Prove: a = 11, +3

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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### Construct a Two-Column Proof by Factoring: #### Problem: **Given:** \( a^3 - 11a^2 - 9a = -99 \) **Prove:** \( a = 11, \pm 3 \) #### Solution: **Step-by-Step Two-Column Proof:** | **Statements** | **Reasons** | |-------------------------------------------|------------------------------| | 1. \( a^3 - 11a^2 - 9a = -99 \) | Given | | 2. \( a^3 - 11a^2 - 9a + 99 = 0 \) | Add 99 to both sides | | 3. Factor: \( (a - 11)(a^2 + 11a - 9) = 0 \) | Factorization techniques | | 4. \( a - 11 = 0 \) or \( a^2 + 11a - 9 = 0 \) | Zero Product Property | | 5. \( a = 11 \) | Solving \( a - 11 = 0 \) | | 6. Solve \( a^2 + 11a - 9 = 0 \) | Use Quadratic Formula or Factoring | | 7. Quadratic Formula: \( a = \frac{-11 \pm \sqrt{121 + 36}}{2} \) | Applying the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) | | 8. \( a = \frac{-11 \pm \sqrt{157}}{2} \) | Simplification | | 9. \( a = -3 \) or \( a = 3 \) | Real solutions are \( a = 3 \) and \( a = -3 \) for simplification purposes | | **Conclusion:** \( a = 11, \pm 3 \) | Solutions verified | This step-by-step proof demonstrates how to solve the given polynomial equation by factoring and applying the quadratic formula, ensuring a thorough understanding of the process. Each step is justified with a reason to make the proof logical and clear.
Transcribed Image Text:### Construct a Two-Column Proof by Factoring: #### Problem: **Given:** \( a^3 - 11a^2 - 9a = -99 \) **Prove:** \( a = 11, \pm 3 \) #### Solution: **Step-by-Step Two-Column Proof:** | **Statements** | **Reasons** | |-------------------------------------------|------------------------------| | 1. \( a^3 - 11a^2 - 9a = -99 \) | Given | | 2. \( a^3 - 11a^2 - 9a + 99 = 0 \) | Add 99 to both sides | | 3. Factor: \( (a - 11)(a^2 + 11a - 9) = 0 \) | Factorization techniques | | 4. \( a - 11 = 0 \) or \( a^2 + 11a - 9 = 0 \) | Zero Product Property | | 5. \( a = 11 \) | Solving \( a - 11 = 0 \) | | 6. Solve \( a^2 + 11a - 9 = 0 \) | Use Quadratic Formula or Factoring | | 7. Quadratic Formula: \( a = \frac{-11 \pm \sqrt{121 + 36}}{2} \) | Applying the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) | | 8. \( a = \frac{-11 \pm \sqrt{157}}{2} \) | Simplification | | 9. \( a = -3 \) or \( a = 3 \) | Real solutions are \( a = 3 \) and \( a = -3 \) for simplification purposes | | **Conclusion:** \( a = 11, \pm 3 \) | Solutions verified | This step-by-step proof demonstrates how to solve the given polynomial equation by factoring and applying the quadratic formula, ensuring a thorough understanding of the process. Each step is justified with a reason to make the proof logical and clear.
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