Question

X2-8x+12=0 (r:2,6) resolva a equação

Answer 1

$x^{2} -8x + 12= 0\\\\ a = 1~~;~~ b= -8~~;~~c=12\\\\\\ \Delta=b^2-4ac\to \Delta=(-8)^2-4.1.12\to \Delta=64- 48\to \boxed{\Delta=16}\\\\ x' \neq x''\\\\ \\ x= \dfrac{-b\pm \sqrt{\Delta} }{2a} \to~~~ x= \dfrac{-(-8)\pm \sqrt{16} }{2.1} \to~~~ x= \dfrac{8\pm 4 }{2} \to\\\\\\ x'= \dfrac{8+ 4 }{2} \to~~~ x'= \dfrac{12 }{2} \to~~~ \boxed{x'=6}\\\\\\ x''= \dfrac{8- 4 }{2} \to~~~ x''= \dfrac{4 }{2} \to~~~ \boxed{x''=2}\\\\\\\\ \large\boxed{\boxed{S=\{2~;~6\}}}$

Answer 2

x² - 8x + 12 = 0

A= 1

B= - 8

C= 12

∆= b² - 4ac

∆= ( - 8)² - 4 • 1 • 12

∆= 64 - 48

∆= 16

x= - b ± √∆ / 2a

x= - ( - 8) ± √16 / 2 • 1

x= 8 ± 4/2

x'= 8 + 4/2 = 12/2 = 6

x''= 8 - 4/2 = 4/2 = 2

S= ( 6 , 2)