Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

in the diagram AB bisects FAE. BF=5x and BE=x2+6. Solve for x.

Sagot :

[tex] If\ AB^{\to}\ bisects\ \angle FAE\ then\ m\angle BAF=m\angle BAE\\\\5x=x^2+6\\\\x^2-5x+6=0\\\\x^2-2x-3x+6=0\\\\x(x-2)-3(x-2)=0\\\\(x-2)(x-3)=0\iff x-2=0\ or\ x-3=0\\\\\boxed{x=2\ or\ x=3}[/tex]