Question

Ayudenme en estas ecuaciones cuadraticas

Answer 1

$3\left(x-1\right)^2+2=0$
$3x^2-6x+5=0$
$x=\frac{-\left(-6\right)+\sqrt{\left(-6\right)^2-4\cdot \:3\cdot \:5}}{2\cdot \:3}=1+i\sqrt{\frac{2}{3}}$
$x=\frac{-\left(-6\right)-\sqrt{\left(-6\right)^2-4\cdot \:3\cdot \:5}}{2\cdot \:3}=1-i\sqrt{\frac{2}{3}}$
$x=1+i\sqrt{\frac{2}{3}},\:x=1-i\sqrt{\frac{2}{3}}$


$x^2-8x+18=0$
$x=\frac{-\left(-8\right)+\sqrt{\left(-8\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}=4+\sqrt{2}i$
$x=\frac{-\left(-8\right)-\sqrt{\left(-8\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}=4-\sqrt{2}i$
$x=4+\sqrt{2}i,\:x=4-\sqrt{2}i$


$\left(x-4\right)\left(x-2\right)=0$
$x-4=0$
$x-4+4=0+4$
$x=4$

$x-2=0$
$x-2+2=0+2$
$x=2$


$\frac{x+\frac{1}{3}}{\frac{1}{3}x+1}=\frac{10x-2}{5x+3}$
$\left(x+\frac{1}{3}\right)\left(5x+3\right)=\left(\frac{1}{3}x+1\right)\left(10x-2\right)$
$x=\frac{9}{5},\:x=1$
$x=\frac{9}{5},\:x=1$