Question

7- Resolva as equações do segundo grau abaixo:
a) x 2 + 3x − 10 = 0
b) 6x 2 + 18x = 0
c) x 2 − 9 = 0
d) 2x 2 − 9x + 7 = 0
e) 4x 2 − 4x + 1 = 0

Answer 1

Oie, Td Bom?!

a)

$x {}^{2} + 3x - 10 = 0$

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$a = 1 \: ,\: b = 3 \:, \: c = - 10$

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$x = \frac{ - b± \sqrt{b {}^{2} - 4ac} }{2a}$

$x = \frac{ - 3± \sqrt{3 {}^{2} - 4 \: . \: 1 \: . \: ( - 10) } }{2 \: . \: 1}$

$x = \frac{ - 3± \sqrt{9 + 40} }{2}$

$x = \frac{ - 3± \sqrt{49} }{2}$

$x = \frac{ - 3±7}{2}$

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$x = \frac{ - 3 + 7}{2} = \frac{4}{2} = 2$

$x = \frac{ - 3 - 7}{2} = \frac{ - 10}{2} = - 5$

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$S = \left \{ x_{1} = - 5 \: ,\: x_{2} = 2 \right \}$

b)

$6x {}^{2} + 18x = 0$

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$a = 6 \: ,\: b = 18 \: , \: c = 0$

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$x = \frac{ - b± \sqrt{b {}^{2} - 4ac} }{2a}$

$x = \frac{ - 18± \sqrt{18 {}^{2} - 4 \: . \: 6 \: . \: 0} }{2 \: . \: 6}$

$x = \frac{ - 8± \sqrt{18 {}^{2} - 0} }{12}$

$x = \frac{ - 8± \sqrt{18 {}^{2} } }{12}$

$x = \frac{ - 18±18}{12}$

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$x = \frac{ - 18 + 18}{12} = \frac{0}{12} = 0$

$x = \frac{ - 18 - 18}{12} = \frac{ - 36}{12} = - 3$

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$S= \left \{ x_{1} = - 3 \: ,\: x_{2} = 0 \right \}$

c)

$x {}^{2} - 9 = 0$

$x {}^{2} = 9$

$x = ± \sqrt{9}$

$x = ±3$

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$S= \left \{ x_{1} = - 3 \: ,\: x_{2} = 3 \right \}$

d)

$2x {}^{2} - 9x + 7 = 0$

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$a = 2 \: , \: b = - 9 \: , \: c = 7$

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$x = \frac{ - b± \sqrt{b {}^{2} - 4ac} }{2a}$

$x = \frac{ - ( - 9)± \sqrt{( - 9) {}^{2} - 4 \: . \: 2 \: . \: 7} }{2 \: . \: 2}$

$x= \frac{9± \sqrt{81 - 56} }{4}$

$x = \frac{9± \sqrt{25} }{4}$

$x = \frac{9±5}{4}$

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$x = \frac{9 + 5}{4} = \frac{14}{4} = \frac{14 \div 2}{4 \div 2} = \frac{7}{2}$

$x = \frac{9 - 5}{4} = \frac{4}{4} = 1$

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$S= \left \{ x_{1} = 1 \: ,\: x_{2} = \frac{7}{2} \right \}$

e)

$4x {}^{2} - 4x + 1 = 0$

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$a = 4 \: , \: b = - 4 \: , \: c = 1$

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$x = \frac{ - b± \sqrt{b {}^{2} - 4ac} }{2a}$

$x = \frac{ - ( - 4)± \sqrt{( - 4){}^{2} - 4 \: . \: 4 \: . \: 1} }{2 \: . \: 4}$

$x = \frac{4± \sqrt{16 - 16} }{8}$

$x = \frac{4± \sqrt{0} }{8}$

$x = \frac{4±0}{8}$

$x = \frac{4}{8} = \frac{4 \div 4}{8 \div 4} = \frac{1}{2}$

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$S= \left \{ x = \frac{1}{2} \right \}$

Att. Makaveli1996