Question

A razão entre as raízes da seguinte equação : 2x ao quadrado +9x-18=0 é : a)-0,25 b)0,50 c)-1,5 d)0,75 e)0,33

Answer 1

Razão 1 :

$\dfrac{x_1}{x_2} = \dfrac{-b +\sqrt{\Delta}}{-b-\sqrt{\Delta}}$

$\dfrac{x_1}{x_2} = \dfrac{b -\sqrt{\Delta}}{b+\sqrt{\Delta}}$

$\dfrac{x_1}{x_2} = \dfrac{b -\sqrt{\Delta}}{b+\sqrt{\Delta}}\cdot \left(\dfrac{1}{1}\right)$

$\dfrac{x_1}{x_2} = \dfrac{b -\sqrt{\Delta}}{b+\sqrt{\Delta}}\cdot \left(\dfrac{b +\sqrt{\Delta}}{b +\sqrt{\Delta}}\right)$

$\dfrac{x_1}{x_2} = \dfrac{b^2 -\Delta}{\left(b+\sqrt{\Delta}\right)^2}$

$\dfrac{x_1}{x_2} = \dfrac{b^2 -\left(b^2-4ac\right)}{\left(b+\sqrt{\Delta}\right)^2}$

$\dfrac{x_1}{x_2} = \dfrac{4ac}{\left(b+\sqrt{\Delta}\right)^2}$

Substituindo:

$\dfrac{x_1}{x_2} = \dfrac{4\cdot2\cdot(-18)}{\left(9+\sqrt{9^2-4\cdot2\cdot(-18)}\right)^2}$

$\dfrac{x_1}{x_2} = \dfrac{4\cdot2\cdot(-18)}{\left(9+\sqrt{225}\right)^2}$

$\dfrac{x_1}{x_2} = \dfrac{4\cdot2\cdot(-18)}{\left(9+15\right)^2}$

$\dfrac{x_1}{x_2} = \dfrac{4\cdot2\cdot(-18)}{\left(24\right)^2}$

$\dfrac{x_1}{x_2} = -\dfrac{1}{4}$

Razão 2 (inverso da razão 1) :

$\dfrac{x_2}{x_1} = -4$

Resposta:

Como a razão 2 não consta no gabarito, temos que a alternativa correta é -0.25 ou -1/4.

a)-0,25