Question

Achar as raízes das equações:
a) x2
- x - 20 = 0

b) x2
- 3x -4 = 0

c) x2
- 8x + 7 = 0

d) x2 - 5x + 6 = 0

precisso disso urgentemente, para entregar hoje!!!!

Answer 1

a)

$x {}^{2} - x - 20 = 0$

$\boxed{a = 1 \:, \: b = - 1 \: , \: c = - 20}$

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

$x = \frac{ - ( - 1)± \sqrt{( - 1) {}^{2} - 4 \: . \: 1 \: . \: ( - 20)} }{2 \: . \: 1}$

$x = \frac{1± \sqrt{1 + 80} }{2}$

$x = \frac{1± \sqrt{81} }{2}$

$x = \frac{1±9}{2}$

$x = \frac{1 + 9}{2} = \frac{10}{2} = \boxed{5}$

$x = \frac{ 1- 9}{2} = \frac{ - 8}{2} = \boxed{ - 4}$

$\boxed{S = \left \{ - 4\: , \:5\right \}}$

b)

$x {}^{2} - 3x - 4 = 0$

$\boxed{a = 1 \: ,\: b = - 3 \:, \: c = - 4}$

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

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

$x = \frac{3± \sqrt{9 + 16} }{2}$

$x = \frac{3± \sqrt{25} }{2}$

$x = \frac{3±5}{2}$

$x = \frac{3 + 5}{2} = \frac{8}{2} = \boxed{4}$

$x = \frac{3 - 5}{2} = \frac{ - 2}{2} = \boxed{ - 1}$

$\boxed{S = \left \{ - 1\: , \:4\right \}}$

c)

$x {}^{2} - 8x + 7 = 0$

$\boxed{a = 1 \:, \: b = - 8 \: ,\: c = 7}$

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

$x = \frac{ - ( - 8)± \sqrt{( - 8) {}^{2} - 4 \: . \: 1 \: . \: 7 } }{2 \: . \: 1}$

$x = \frac{8± \sqrt{64 - 28} }{2}$

$x = \frac{8± \sqrt{36} }{2}$

$x = \frac{8±6}{2}$

$x = \frac{8 + 6}{2} = \frac{14}{2} = \boxed{7}$

$x = \frac{8 - 6}{2} = \frac{2}{2} = \boxed{1}$

$\boxed{S = \left \{1\: , \:7\right \}}$

d)

$x {}^{2} - 5x + 6 = 0$

$\boxed{a = 1 \: ,\: b = - 5 \:, \: c = 6}$

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

$x = \frac{ - ( - 5)± \sqrt{( - 5) {}^{2} - 4 \: . \: 1 \: . \: 6} }{2 \: . \: 1}$

$x = \frac{5± \sqrt{25 - 24} }{2}$

$x = \frac{5± \sqrt{1} }{2}$

$x = \frac{5±1}{2}$

$x = \frac{5 + 1}{2} = \frac{6}{2} = \boxed{3}$

$x = \frac{5 - 1}{2} = \frac{4}{2} = \boxed{2}$

$\boxed{S = \left \{2\: , \:3\right \}}$

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