Question

quais as raízes da seguinte equação do 2° grau : x² - 5x + 4 = 0 ?

Answer 1

$x^2-5x+4=0$

$x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-\left(-5\right)\pm \:3}{2\cdot \:1}$

$x_1=\frac{-\left(-5\right)+3}{2\cdot \:1},\:x_2=\frac{-\left(-5\right)-3}{2\cdot \:1}$

$\frac{-\left(-5\right)+3}{2\cdot \:1}\\\\=\frac{5+3}{2\cdot \:1}\\\\=\frac{8}{2\cdot \:1}\\\\=\frac{8}{2}\\\\=4$

$\frac{-\left(-5\right)-3}{2\cdot \:1}$

$=\frac{5-3}{2\cdot \:1}\\\\=\frac{2}{2\cdot \:1}\\\\=\frac{2}{2}\\\\=1$

$x=4,\:x=1$

Answer 2

Resposta:

x'= 4 x"=4

$x'= \frac{-(-5) + \sqrt{(-5)^{2}- 4.1.4 }}{2.1}$

$x'= \frac{-(-5) - \sqrt{(-5)^{2}- 4.1.4 }}{2.1}$