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Understanding Theorem 10.16: Finding Lengths through Quadratic Equations

This instructional content guides learners through the application of Theorem 10.16 to find lengths in geometric contexts. Using quadratic equations and the quadratic formula, students will practice substituting values and simplifying equations to determine the positive solutions for lengths, which cannot be negative. The content includes several examples and guided practice problems, assisting in mastering the theorem's practical use. Each example reinforces the principles of simplification and standard form application while ensuring clarity in the solution process.

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Understanding Theorem 10.16: Finding Lengths through Quadratic Equations

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  1. RQ2 = RS RT 162 = x (x + 8) 256 = x2 + 8x 0 = x2 + 8x – 256 82 – 4(1) (– 256) –8 + x = 2(1) 17 x = – 4 + 4 EXAMPLE 3 Find lengths using Theorem 10.16 Use the figure at the right to find RS. SOLUTION Use Theorem 10.16. Substitute. Simplify. Write in standard form. Use quadratic formula. Simplify.

  2. So, x 12.49, andRS 12.49 17 = – 4 + 4 EXAMPLE 3 Find lengths using Theorem 10.16 Use the positive solution, because lengths cannot be negative.

  3. x2 x2 = 4 = 1 (1 + 3) x = 2 for Example 3 GUIDED PRACTICE Find the value of x. 4. SOLUTION Use Theorem 10.16. Simplify. Simplify.

  4. 72 49 = 25 + 5x = 5 (x + 5) = 5x 24 x = 24 5 for Example 3 GUIDED PRACTICE Find the value of x. 5. SOLUTION Use Theorem 10.16. Simplify. Write in standard form. Simplify.

  5. 122 144 = x2 + 10x = x (x + 10) 0 = x2 + 10x – 144 x = 102 – 4(1) (– 144) –10 + x = 8 2(1) for Example 3 GUIDED PRACTICE Find the value of x. 6. SOLUTION Use Theorem 10.16. Simplify. Write in standard form. Use quadratic formula. Simplify.

  6. 152 = x (x + 14) 255 = x2 + 14x 0 = x2 + 14x – 255 x 142 – 4(1) (– 255) –14 + = 2(1) x = – 7 + 274 for Example 3 GUIDED PRACTICE 7. Determine which theorem you would use to find x. Then find the value of x. SOLUTION Theorem 10.16 Use Theorem 10.16. Simplify. Write in standard form. Use quadratic formula. Simplify.

  7. x (18) = (9) (16) 18x = 144 x = 8 for Example 3 GUIDED PRACTICE 8. Determine which theorem you would use to find x. Then find the value of x. SOLUTION Use Theorem 10.14. Substitute. Simplify. Simplify.

  8. 18 (18 + 22) = 29x + x2 720 0 = x2 + 29x – 720 = x (x + 29) x 292 – 4(1) (– 720) –29 + = 2(1) x = 16 for Example 3 GUIDED PRACTICE 8. Determine which theorem you would use to find x. Then find the value of x. SOLUTION Use Theorem 10.15. Substitute. Simplify. Write in standard form. Use quadratic formula. Simplify.

  9. 9. In the diagram for Theorem 10.16, what must be true about ECcompared toEA ? for Example 3 GUIDED PRACTICE SOLUTION EC < EA

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