Question

Sabendo que x-1, x+5, e 11x-1 formam, nessa ordem , uma PG , determine esses tres termos

Answer 1

$\dfrac{11x-1}{x+5}=\dfrac{x+5}{x-1}=(x+5)^{2}=(x-1)(11x-1)\Rightarrow\\ \\ \\ x^{2}+10x+25=11x^{2}-x-11x+1\Rightarrow 10x^{2}-22x-24=0\Rightarrow\\ \\ \\ 10x^{2}-22x-24=0 \div 2\ (dividimos\ por\ 2\ para\ facilitar\ o\ calculo):\\ \\ 5x^{2}-11x-12=0\\ \\ x=\dfrac{-b\pm\sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}\Rightarrow x=\dfrac{11\pm\sqrt{\left(-11\right)^{2}-4\cdot 5\cdot \left(-12\right)}}{2\cdot 5}\Rightarrow\\ \\ \\ x=\dfrac{11\pm\sqrt{121+240}}{10}\Rightarrow x=\dfrac{11\pm 19}{10}\Rightarrow$


$x_{1}=\dfrac{11+19}{10}\Rightarrow x_{1}=3\\ \\ \\ x_{2}=\dfrac{11-19}{10}\Rightarrow x_{2}=\dfrac{-8}{10}\Rightarrow x_{2}=\dfrac{-4}{5}$


Portanto a PG é:

Para x = 3 ⇒ (x-1, x+5, 11x-1) ⇒ (3-1, 3+5, 11·3 - 1) ⇒ (2, 8, 32)

Para x = -4/5 ⇒

$x-1=-\dfrac{4}{5}-1\Rightarrow \dfrac{-4-5}{5}\Rightarrow -\dfrac{9}{5}\\ \\ \\ x+5=-\dfrac{4}{5}+5\Rightarrow \dfrac{-4+25}{5}\Rightarrow \dfrac{21}{5}\\ \\ \\ 11x-1=11\cdot \left(-\dfrac{4}{5}\right)-1\Rightarrow -\dfrac{44}{5}-1\Rightarrow \dfrac{-44-5}{5}\Rightarrow \dfrac{-49}{5}$

$\left(-\dfrac{9}{5},\dfrac{21}{5},-\dfrac{49}{5}\right)$