Question

determine o valor de x na progressao (x-1,x+2,3x,)e escreva a pg.​

Answer 1

$\Large\boxed{\begin{array}{l}\sf (x-1,x+2,3x)\\\sf 3x\cdot(x-1)=(x+2)^2\\\sf 3x^2-3x=x^2+4x+4\\\sf 3x^2-x^2-3x-4x-4=0\\\sf 2x^2-7x-4=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-7)^2-4\cdot2\cdot(-4)\\\sf\Delta=49+32\\\sf\Delta=81\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-(-7)\pm\sqrt{81}}{2\cdot2}\\\\\sf x=\dfrac{7\pm9}{4}\begin{cases}\sf x_1=\dfrac{7+9}{4}=\dfrac{16}{4}=4\\\\\sf x_2=\dfrac{7-9}{4}=-\dfrac{2}{4}=-\dfrac{1}{2}\end{cases}\end{array}}$

$\Large\boxed{\begin{array}{l}\sf para~x=4:\\\sf (4-1,4+2,3\cdot4)=(3,6,12)\\\sf para~x=-\dfrac{1}{2}:\\\sf \bigg(-\dfrac{1}{2}-1,-\dfrac{1}{2}+2,3\cdot\bigg(-\dfrac{1}{2}\bigg)\bigg)\\\\\sf\bigg(-\dfrac{3}{2},\dfrac{3}{2},-\dfrac{3}{2}\bigg)\end{array}}$