Question

Qual a razão da P.G. (x, 2x+4, 10x-4,...) ? Por favor me ajudem!

Answer 1

Resposta:

q= -4 ou q=3

Explicação passo-a-passo:

PG:

$\displaystyle q=\frac{2x+4}{x} =\frac{10x-4}{2x+4} \\\\\\(2x+4)^2=x(10x-4)\\4x^2+16x+16=10x^2-4x\\(10-4)x^2-(4+16)x-16=0\\6x^2-20x-16=0~~(\div2)\\3x^2-10x-8=0$

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~3x^{2}-10x-8=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=3{;}~b=-10~e~c=-8\\\\\Delta=(b)^{2}-4(a)(c)=(-10)^{2}-4(3)(-8)=100-(-96)=196\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-10)-\sqrt{196}}{2(3)}=\frac{10-14}{6}=\frac{-4\div2}{6\div2}=-\frac{2}{3} \\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-10)+\sqrt{196}}{2(3)}=\frac{10+14}{6}=\frac{24}{6}=4\\\\S=\{-\frac{2}{3},~4\}$Para x= -2/3

$\displaystyle q=\frac{2(-\frac{2}{3}) +4}{-\frac{2}{3} } =\frac{-\frac{4}{3}+4 }{-\frac{2}{3}} =\frac{\frac{12-4}{3} }{-\frac{2}{3}} =\frac{\frac{8}{3} }{-\frac{2}{3}} =\frac{8}{3}.\frac{-3}{2} =-4$

Para x=4

$\displaystyle q=\frac{2(4) +4}{4} =\frac{8+4 }{4} =\frac{12}{4} =3$