Question

При якому додатьньому значенні х числа 3х-2, х+2, х+8 будуть послідовними членами геометричної прогресії? ​

Answer 1

$\displaystyle\bf\\b_{1} =3x-2\\\\b_{2} =x+2\\\\b_{3} =x+8\\\\\\b_{3} :b_{2} =b_{2} :b_{1} \\\\\\\frac{x+8}{x+2} =\frac{x+2}{3x-2} \\\\\\(x+8)\cdot(3x-2)=(x+2)\cdot(x+2)\\\\\\3x^{2} -2x+24x-16=x^{2} +4x+4\\\\\\3x^{2} -2x+24x-16-x^{2} -4x-4=0\\\\\\2x^{2} +18x-20=0 \ |:2\\\\\\x^{2} +9x-10=0\\\\\\Teorema \ Vieta:\\\\\\x_{1} +x_{2} =-9\\\\x_{1} \cdot x_{2} =-10\\\\\\\boxed{x_{1}=-10 \ \ , \ \ x_{2} =1 }$