Question

na figura, temos PA= x, PB=6, PC= x+1 e PD= x+2. Determine a medida dos seguintes segmentos PA, PC e PD.

Answer 1

$\boxed{\begin{array}{l}\sf PA\cdot PB= PC\cdot PD\\\sf x\cdot6=(x+1)\cdot(x+2)\\\sf x^2+3x+2=6x\\\sf x^2+3x-6x+2=0\\\sf x^2-3x+2=0\\\sf \Delta=b^2-4ac\\\sf\Delta=(-3)^2-4\cdot1\cdot2\\\sf\Delta=9-8\\\sf\Delta=1\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-(-3)\pm\sqrt{1}}{2\cdot1}\\\\\sf x=\dfrac{3\pm1}{2}\begin{cases}\sf x_1=\dfrac{3+1}{2}=\dfrac{4}{2}=2\\\\\sf x_2=\dfrac{3-1}{2}=\dfrac{2}{2}=1\end{cases}\\\\\sf PA=2~ou~1\\\sf PC=3~ou~2\\\sf PD=4~ou~3\end{array}}$