Question

preciso de ajudar o mais rápido possível PFV !!!

Answer 1

Olá...

teorema de Talles

$\frac{AD}{BD} = \frac{AE}{CE} \\ \\ \frac{x-1}{3} = \frac{x+4}{x} \\ \\ x(x-1)=3(x+4) \\ \\ x^{2} -x=3x+12 \\ x^{2} -x-3x-12=0 \\ x^{2} -4x-12=0 \\ \\ a=1~~~~b=-4~~~~c=-12 \\ \\ \Delta=b^2-4ac \\ \Delta=(-4)^2-4(1)(-12) \\ \Delta=16+48 \\ \Delta=64$ 

$x= \frac{-b\pm \sqrt{\Delta} }{2a} = \frac{-(-4)\pm \sqrt{64} }{2} = \frac{4\pm8}{2 } \\ \\ x'= \frac{4+8}{2} = \frac{12}{2} =6 \\ \\ x"= \frac{4-8}{2} =- \frac{4}{2} =-2~~n/~serve$ 

$AB=x-1+3 \\ AB=6-1+3 \\ AB=8 \\ \\ AC=x+4+x \\ AC=6+4+6 \\ AC=16$