Resposta :

Phellyppe

É só resolver utilizando a fórmula de Bháskara:

 

x²-3x+2=0

a=1

b=3

c=2

 

∆ = b² – 4ac  =>  ∆ = (-3)² - 4*1*2  => ∆ = 9 - 8  => ∆ = 1

 

x= -b +/- √∆ /2a

x = -(-3) +/- √1 / 2*1

x = 3 +/- 1 /2

 

x' = 3 + 1 / 2 => x'= 4/2 => x' =2

 

x'' = 3 - 1 / 2  => x'' = 2/2  => x'' = 1

 

PeH

[tex]\blacktriangleright x^2 - 3x + 2 = 0 \\\\ x ={-b\pm\sqrt{b^2-4ac} \over 2a} \\\\ x ={3\pm\sqrt{9 - 4 \cdot 1 \cdot 2} \over 2} \\\\ x ={3\pm\sqrt{9 - 8} \over 2} \\\\ x = \frac{3 \pm 1}{2} \rightarrow x_1 = 2 \ \text{e} \ x_2 = 1[/tex]

 

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