Question

Determine as raizes ou zeros da equação do segundo grau abaixo.
x2- 6x +5=0​

Answer 1

Resposta:

Explicação passo-a-passo:

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-6x+5=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=1{;}~b=-6~e~c=5\\\\C\'alculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(-6)^{2}-4(1)(5)=36-(20)=16\\\\C\'alculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-6)-\sqrt{16}}{2(1)}=\frac{6-4}{2}=\frac{2}{2}=1\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-6)+\sqrt{16}}{2(1)}=\frac{6+4}{2}=\frac{10}{2}=5\\\\S=\{1,~5\}$

Answer 2

Resposta:

A)

y = x² - 6x + 5

a = 1    b = -6    c = 5

delta = b² - 4 . a . c

delta = -6² - 4 .1 . 5

delta = 36 - 20

delta = 16

x = -b +- raiz de delta / 2 . a

x = - (-6) +- raiz de 16 / 2 . 1

x¹= 6 + 4 / 2

x¹=5

x²=6 - 4 / 2

x²= 2