Question

determine x, de modo que a sequência (4,4x,10x+6) seja pg

Answer 1

Numa $PG$, $\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}$.

$a_1=4$, $a_2=4x$ e $a_3=10x+6$:

$\dfrac{4x}{4}=\dfrac{10x+6}{4x}$

$x=\dfrac{10x+6}{4x}$

$4x^2=10x+6$

$4x^2-10x-6=0$

$2x^2-5x-3=0$

$\Delta=(-5)^2-4\cdot2\cdot(-3)=25+24=49$

$x=\dfrac{-(-5)\pm\sqrt{49}}{2\cdot2}=\dfrac{5\pm7}{4}$

$x'=\dfrac{5+7}{4}=\dfrac{12}{4}=3$;

$x"=\dfrac{5-7}{4}=\dfrac{-2}{4}=-\dfrac{1}{2}$.