itsandypandy
Answered

Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

Solve this for Brainliest!

See the attachment [Level - GCSE]

Solve This For BrainliestSee The Attachment Level GCSE class=

Sagot :

gvimalaratnam21

Answer:

[tex]x = 3 + 4\sqrt{3}[/tex]

Step-by-step explanation:

OPQ is a right angle triangle.

Using Pythagoras

[tex]x^{2} + (x+5)^{2} = (x+8)^{2}\\x^{2} + x^{2} + 10x +25 = x^{2} + 16x + 64\\x^{2} + x^{2} - x^{2} + 10x - 16x + 25 - 64 = 0\\x^{2} - 6x -39 = 0[/tex]

Using quadratic formula

[tex]x = \frac{-b \± \sqrt{b^{2}-4ac}}{2a}\\x = \frac{-(-6) \± \sqrt{(-6)^{2}-4(1)(-39)}}{2(1)}\\x = \frac{6}{2} \± \frac{\sqrt{192}}{2}\\x = 3 \± \frac{8\sqrt{3}}{2}\\x = 3 \± 4\sqrt{3}[/tex]

[tex]x = 3 + 4\sqrt{3} = 9.928(4s.f.)[/tex]

             OR

[tex]x = 3 - 4\sqrt{3} = -3.928(4s.f.)[/tex] (Length can't be negative)

∴ [tex]x = 3 + 4\sqrt{3}[/tex]

vgowthaman24

Answer:

[tex]x=3+4\sqrt{3}[/tex]

Step-by-step explanation:

I agree with the another person's steps.

We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.