Question

ME AJUDEEM PFVR......​

Answer 1

Resposta:

A) $x=2,\:x=\frac{1}{3}$

B) $x=3$

Explicação passo-a-passo:

A)

$\mathrm{Para\:}\quad a=3,\:b=-7,\:c=2:\quad x_{1,\:2}=\frac{-\left(-7\right)\pm \sqrt{\left(-7\right)^2-4\cdot \:3\cdot \:2}}{2\cdot \:3}\\\\\frac{-\left(-7\right)+\sqrt{\left(-7\right)^2-4\cdot \:3\cdot \:2}}{2\cdot \:3}\\\\\mathrm{Aplicar\:a\:regra}\:-\left(-a\right)=a\\\\=\frac{7+\sqrt{\left(-7\right)^2-4\cdot \:3\cdot \:2}}{2\cdot \:3}$

$=\frac{7+\sqrt{25}}{2\cdot \:3}\\\\=\frac{7+\sqrt{25}}{6}\\\\=\frac{7+5}{6}\\\\=\frac{12}{6}\\\\=2$

========================

$\frac{-\left(-7\right)-\sqrt{\left(-7\right)^2-4\cdot \:3\cdot \:2}}{2\cdot \:3}\\\\\mathrm{Aplicar\:a\:regra}\:-\left(-a\right)=a\\\\=\frac{7-\sqrt{\left(-7\right)^2-4\cdot \:3\cdot \:2}}{2\cdot \:3}\\\\=\frac{7-\sqrt{25}}{2\cdot \:3}\\\\=\frac{7-\sqrt{25}}{6}\\\\=\frac{7-5}{6}\\\\=\frac{2}{6}\\\\\mathrm{Eliminar\:o\:fator\:comum:}\:2\\\\=\frac{1}{3}$

====================

B)

$\mathrm{Para\:}\quad a=1,\:b=-6,\:c=9:\quad x_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:1\cdot \:9}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{0}}{2\cdot \:1}\\\\x=\frac{-\left(-6\right)}{2\cdot \:1}\\\\=\frac{6}{2\cdot \:1}\\\\=\frac{6}{2}\\\\\mathrm{Dividir:}\:\frac{6}{2}=3\\\\=3$