Question

resolva as seguintes equações do segundo grau

a) x²-5x +6=0




b) x²-8x+12=0

Answer 1

$A) ~~x^2-5x +6=0\\\\ a=1;~b=-5;~c=6\\\\\\ \Delta=b^2-4ac\to~ \Delta=(-5)^2-4.1.6\to~ \Delta=25-24\to~\boxed{ \Delta=1}\\\\ x' \neq x'' \\\\\\ x= \dfrac{-b \pm \sqrt{ \Delta} }{2a} \to~~ x= \dfrac{-(-5) \pm \sqrt{ 1} }{2.1} \to~~ x= \dfrac{5 \pm 1 }{2} \to\\\\\\ x'= \dfrac{5 +1 }{2} \to~~ x'= \dfrac{6 }{2} \to~~ \boxed{x'=3}\\\\\\ x'= \dfrac{5 -1 }{2} \to~~ x'= \dfrac{4 }{2} \to~~ \boxed{x'=2}\\\\\\\\\\ \large\boxed{S=\{2~;~3\}}$







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

Answer 2

x² - 5x + 6 = 0
a = 1; b = - 5; c = 6

Δ = b² - 4ac
Δ = (-5)² - 4.1.6
Δ = 25 - 24
Δ = 1

x = - b +/- √Δ = - ( - 5) +/- √1
           2a                2.1

x = 5 + 1 = 6/2 = 3
        2

x = 5 - 1 = 4/2 = 2 
        2

R.: x = 3 e x = 2
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b)
x² - 8x + 12 = 0
a = 1; b = - 8; c = 12

Δ = b² - 4ac
Δ = (-8)² - 4.1.12
Δ = 64 - 48
Δ = 16

x = - b +/- √Δ = - ( - 8) +/- √16
           2a                2.1

x = 8 + 4 = 12/2 = 6
         2

x = 8 - 4 = 4/2 = 2
        2

R.: x = 6 e x = 2