Question

05-Resolvendo a equação do 2° grau x? - 8x + 15 = 0, obtemos as raizes:
a) +3,+5
b) -3,+6
c) -3,5
d) -5,-2
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Answer 1

Resposta:

Alternativa a) 3 , 5

Explicação passo-a-passo:

x² - 8 x + 15 = 0

a = 1 b = -8 c = 15

∆ = b² - 4 . a . c

∆ = (-8)² - 4 . 1 . 15

∆ = 64 - 60

∆ = 4

X = -b +/- ✓∆ ÷ 2 . a

X = -(-8) +/- ✓4 ÷ 2 . 1

X = 8 +/- 2 ÷ 2

X' = 8 + 2 / 2 = 10 / 2 = 5

X'' = 8 - 2 /2 = 6 / 2 = 3

Answer 2

Para encontrarmos as raízes de uma equação do 2° grau, aplique a fórmula de Bhaskara

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$\underbrace{Veja:}$

$\sf x^2-8x+15=0$

coeficientes:

• a = 1
• b = -8
• c = 15

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$\sf \Delta=b^2-4ac$

$\sf \Delta=(-8)^2-4.(1).(15)$

$\sf \Delta=64-60$

$\sf \Delta=4$

$\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}$

$\sf x=\dfrac{-(-8)\pm\sqrt{4}}{2.(1)}~~\Rightarrow~~x=\dfrac{8\pm2}{2}$

$\sf x'=\dfrac{8+2}{2}~~\Rightarrow~~x'=\dfrac{10}{2}~~\Rightarrow~~\boxed{\sf x'=5}$

$\sf x''=\dfrac{8-2}{2}~~\Rightarrow~~x''=\dfrac{6}{2}~~\Rightarrow~~\boxed{\sf x''=3}$

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conjunto solução é: $\boxed{\sf S=\left\{3~~;~~5\right\}}$

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Resposta: Letra A