1 / 9

EXAMPLE 3

RQ 2. = RS RT. 16 2. = x ( x + 8). 256. = x 2 + 8 x. 0. = x 2 + 8 x – 256. 8 2 – 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.

Download Presentation

EXAMPLE 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related