Question

Resolva as equações do 2º grau através da fórmula de bhaskara.
a) x2 - 4x + 4 =0
b) x² - 8x + 15 = 0.
c) x2 + 2x = 0
d) 3 x- 3 = 0​

Answer 1

Olá!!  :)

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ITEM (A):

$\sf x^{2} \ - \ 4x \ + \ 4 \ = \ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ \ a \ = \ 1 \ \ \ \ ; \ \ \ \ b \ = \ -4 \ \ \ \ ; \ \ \ \ c \ = \ 4 \\\\ \\\\ \Delta \ = \ b^{2} \ - \ 4ac \\\\ \Delta \ = \ (-4)^{2} \ - \ 4 \ . \ 1 \ . \ 4 \\\\ \Delta \ = \ 16 \ - \ 16 \\\\ \Delta \ = \ 0$

Quando delta vale zero, as duas raízes são iguais.

Portanto, só precisamos calcular uma vez:

$\sf x \ = \ \frac{-b \ \ + \ \ \sqrt{\Delta}}{2a} \\\\ x \ = \ \frac{-(-4) \ \ + \ \ \sqrt{0}}{2 \ . \ 1} \\\\ x \ = \ \frac{4 \ + \ 0}{2 \ . \ 1} \\\\ x \ = \ \frac{4}{2} \\\\ \boxed{x \ = \ 2}$                  

Resposta:

$\sf S \ = \ \{2,2\}$

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ITEM (B):

$\sf x^{2} \ - \ 8x \ + \ 15 \ = \ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow \ \ a \ = \ 1 \ \ \ \ ; \ \ \ \ b \ = \ -8 \ \ \ \ ; \ \ \ \ c \ = \ 15 \\\\ \\ \Delta \ = \ b^{2} \ - \ 4ac \\\\ \Delta \ = \ (-8)^{2} \ - \ 4 \ . \ 1 \ . \ 15 \\\\ \Delta \ = \ 60 \ - \ 60 \\\\ \Delta \ = \ 0$

Mesmo esquema do item anterior:

quando x vale zero, as raízes são iguais e calculamos apenas uma vez.

$\sf x \ = \ \frac{-b \ \ + \ \ \sqrt{\Delta}}{2a} \\\\ x \ = \ \frac{-(-8) \ \ + \ \ \sqrt{0}}{2 \ . \ 1} \\\\ x \ = \ \frac{8 \ + \ 0}{2 \ . \ 1} \\\\ x \ = \ \frac{8}{2} \\\\ \boxed{x \ = \ 4}$

Resposta:

$\sf S \ = \ \{2, \ 2\}$

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ITEM (C):

$\sf x^{2} \ + \ 2x \ = \ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow \ \ a \ = \ 1 \ \ \ \ ; \ \ \ \ b \ = \ 2 \ \ \ \ ; \ \ \ \ c \ = \ 0 \\\\ \\ \Delta \ = \ b^{2} \ - \ 4ac \\\\ \Delta \ = \ 2^{2} \ - \ 4 \ . \ 1 \ . \ 0 \\\\ \Delta \ = \ 4$

$\sf x' \ = \ \frac{-b \ \ + \ \ \sqrt{Delta}}{2a} \\\\ x' \ = \ \frac{-2 \ \ + \ \ \sqrt{4}}{2 \ . \ 1} \\\\ x' \ = \ \frac{-2 \ + \ 2}{2} \\\\ x' \ = \ \frac{0}{2} \\\\ \boxed{x' \ = \ 0}$               $\sf x'' \ = \ \frac{-b \ \ - \ \ \sqrt{\Delta}}{2a} \\\\ x'' \ = \ \frac{-2 \ \ - \ \ \sqrt{4}}{2 \ . \ 1} \\\\ x'' \ = \ \frac{-2 \ - \ 2}{2} \\\\ x'' \ = \ - \ \frac{4}{2} \\\\ \boxed{x'' \ = \ -2}$

Resposta:

$S \ = \ \{0, \ -2\}$

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ITEM (D):

Perceba os termos dessa equação:

$\sf 3x \ - \ 3 \ = \ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ \ a \ = \ 3 \ \ \ \ ; \ \ \ \ b \ = \ 0 \ \ \ \ ; \ \ \ \ c \ = \ -3$

Quando b vale zero, não calculamos fórmula de Bhaskara.

Apenas isolamos x:

$\Rightarrow \ \ 3x \ - \ 3 \ = \ 0 \\\\ \Rightarrow \ \ 3x \ = \ 3 \\\\ \Rightarrow \ \ x \ = \ \frac{3}{3} \\\\ \Rightarrow \ \ x \ = \ 1$

Resposta:

$S \ = \ \{1\}$

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Espero ter ajudado, bons estudos!!  :)