1. EXAMPLE 3 Find lengths using Theorem 10.16 Use the figure at the right to find RS. SOLUTION 256= x 2 + 8x 0= x 2 + 8x – 256 RQ 2 = RS RT 16 2 = x (x + 8) x –8 + 8 2 – 4(1) (– 256) 2(1) = x = – 4 + 4 17 Use Theorem 10.16. Substitute. Simplify. Write in standard form. Use quadratic formula. Simplify.

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

3. GUIDED PRACTICE for Example 3 Find the value of x. SOLUTION x2x2 = 1 (1 + 3) x = 2 Use Theorem 10.16. Simplify. 4. = 4 x2x2

4. GUIDED PRACTICE for Example 3 Find the value of x. SOLUTION 5. 49= 25 + 5x 24 = 5x 7272 = 5 (x + 5) Use Theorem 10.16. Simplify. Write in standard form. Simplify. x = 24 5

5. GUIDED PRACTICE for Example 3 Find the value of x. SOLUTION 6. 144= x 2 + 10x 0= x 2 + 10x – 144 12 2 = x (x + 10) x –10 + 10 2 – 4(1) (– 144) 2(1) = Use Theorem 10.16. Simplify. Write in standard form. Use quadratic formula. Simplify. x = 8

6. GUIDED PRACTICE for Example 3 Determine which theorem you would use to find x. Then find the value of x. 7. Theorem 10.16 SOLUTION 255= x 2 + 14x 0= x 2 + 14x – 255 15 2 = x (x + 14) x –14 + 14 2 – 4(1) (– 255) 2(1) = Use Theorem 10.16. Simplify. Write in standard form. Use quadratic formula. Simplify. x = – 7 + 274

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

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

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