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Mastering Radical Equations: Steps and Examples for Solving with Extraneous Solutions

This guide focuses on solving radical equations, illustrating key concepts and steps required for proper solutions. We define extraneous solutions and provide clear examples to ensure a comprehensive understanding. The steps include isolating the radicand, squaring both sides, and checking each solution's validity. Each example demonstrates the process, identifying valid solutions and those that are extraneous. Learn to confidently tackle radical equations and avoid common pitfalls with this educational resource.

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Mastering Radical Equations: Steps and Examples for Solving with Extraneous Solutions

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  1. Warm-up Factor. 1. p2 + 6p – 27 2. p2 – 15p – 16 3. k2 + 9k + 20 4. k2 – 14k + 24

  2. Warm-up Factor. 1. p2 + 6p – 27 2. p2 – 15p – 16 -27 -16 -3 9 1 -16 6 -15 (p – 3)(p + 9) (p + 1)(p – 16)

  3. Warm-up Factor. 3. k2 + 9k + 20 4. k2 – 14k + 24 20 24 4 5 -2 -12 9 -14 (k + 4)(k + 5) (k – 2)(k – 12)

  4. Section 10.4 Solving Radical Equations (Part 2) Standards: 2.0, 25.2 Objective: I will solve equations containing radicals.

  5. Extraneous Solutions: An answer that does not solve the original equation. x = 2 x2 = 4 (x)2 = 22 x = 2 x = -2

  6. Steps: Get the radicand by itself. Square both sides. Solve the remaining equation. Check your answer and identify extraneous solutions.

  7. Example 1: -6 2 -3 -1 (x + 2)(x – 3) = 0 x + 6 x2 = x + 2 = 0 x – 3 = 0 -x – 6 -x – 6 x2 – x – 6 = 0 x = -2 x = 3

  8. Example 1: check x = -2 & 3 ✔ -2 = 2 3 = 3 extraneous

  9. Example 2: 4 -1 -4 -5 (x – 1)(x – 4) = 0 5x – 4 x2 = x – 1 = 0 x – 4 = 0 -5x +4 -5x + 4 x2 – 5x + 4 = 0 x = 1 x = 4

  10. Example 2: check x = 1 & 4 ✔ ✔ 1 = 1 4 = 4

  11. Example 3: -20 -4 5 1 (x – 4)(x + 5) = 0 20 – x x2 = x – 4 = 0 x + 5 = 0 -20 + x -20 + x x2 + x – 20 = 0 x = 4 x = -5

  12. Example 3: check x = 4 & -5 ✔ 4 = 4 -5 = 5 extraneous

  13. Example 4: -56 -7 8 1 (x – 7)(x + 8) = 0 56 – x x2 = x – 7 = 0 x + 8 = 0 -56 + x -56 + x x2 + x – 56 = 0 x = 7 x = -8

  14. Example 4: check x = 7 & -8 ✔ 7 = 7 -8 = 8 extraneous

  15. Example 5: -12 3 -4 -1 (x + 3)(x – 4) = 0 x + 12 x2 = x + 3 = 0 x – 4 = 0 -x – 12 -x – 12 x2 – x – 12 = 0 x = -3 x = 4

  16. Example 5: check x = -3 & 4 ✔ -3 = 3 4 = 4 extraneous

  17. Example 6: -15 3 -5 -2 (x + 3)(x – 5) = 0 2x + 15 x2 = x + 3 = 0 x – 5 = 0 -2x – 15 -2x – 15 x2 – 2x – 15 = 0 x = -3 x = 5

  18. Example 6: check x = -3 & 5 ✔ -3 = 3 5 = 5 extraneous

  19. Example 7: 8 2 4 6 (x + 2)(x + 4) = 0 -8 – 6x x2 = x + 2 = 0 x + 4 = 0 8 + 6x 8 + 6x x2 + 6x + 8 = 0 x = -2 x = -4

  20. Example 7: check x = -2 & -4 -2 = 2 -4 = 4 extraneous extraneous no solution

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