Question

Reduza à forma normal as seguintes equações do 2.° grau:

Resolver as seguintes equações do 2.° grau, sendo U = R:

A) x2 - 7 = 10x + 4
B) 2x2 = 5x - 8
C) 3x2 - x = 2x2 + 2
D) x2 + 4x = 1 - x12
E) x2 - x = x - 1
F) 10x2 - 1 = 3x
G) x2 = 3x + 4
H) 9x2 - 5x = x - 1​

Answer 1

$a) \\ x {}^{2} - 7 = 10x + 4 \\ x {}^{2} - 7 - 10x - 4 = 0 \\ x {}^{2} - 10x - 11 = 0 \\ \boxed{a = 1 \:, \: b = - 10 \: , \: c = - 11} \\ x = \frac{ - b ±\sqrt{b {}^{2} - 4ac} }{2a} \\ x = \frac{ - ( - 10)± \sqrt{( - 10) {}^{2} - 4 \: . \: 1 \: . \: ( - 11)} }{2 \: . \: 1} \\ x = \frac{10± \sqrt{100 + 44} }{2} \\ x = \frac{10± \sqrt{144} }{2} \\ x = \frac{10± 12}{2} \\ x = \frac{10 + 12}{2} = \frac{22}{2} = \boxed{\boxed{ \boxed{11}}} \\ x = \frac{10 - 12}{2} = \frac{ - 2}{2} = \boxed{\boxed{ \boxed{ - 1}}}$

$b) \\ 2x {}^{2} = 5x - 8 \\ 2x {}^{2} - 5x - 8 = 0 \\ \boxed{a = 2 \: ,\: b = - 5 \: , \: c = - 8} \\ x = \frac{ - b ±\sqrt{b {}^{2} - 4ac } }{2a} \\ x = \frac{ - ( - 5) ±\sqrt{( - 5) {}^{2} - 4 \: . \: 2 \: . \: 8} }{2 \: . \: 2} \\ x = \frac{5 ± \sqrt{25 - 64} }{4} \\ x = \frac{5 ± \sqrt{ - 39} }{4} \\ \boxed{\boxed{\boxed{x∉ℝ}}}$

$c) \\ 3x {}^{2} - x = 2x {}^{2} + 2 \\ 3x {}^{2} - x - 2x {}^{2} - 2 = 0 \\ x {}^{2} - x - 2 = 0 \\ \boxed{a = 1 \: ,\: b = - 1 \: ,\: c = - 2} \\ x = \frac{ - b ±\sqrt{b {}^{2} - 4ac} }{2a} \\ x = \frac{ - ( - 1)± \sqrt{( - 1) {}^{2} - 4 \: . \: 1 \: . \: ( - 2)} }{2 \: . \: 1} \\ x = \frac{1 ±\sqrt{1 + 8} }{2} \\ x = \frac{1 ±\sqrt{9} }{2} \\ x = \frac{1 ±3}{2} = \\ x = \frac{1 + 3}{2} = \frac{4}{2} = \boxed{\boxed{\boxed{2}}} \\ x = \frac{1 - 3}{2} = \frac{ - 2}{2} = \boxed{\boxed{\boxed{ - 1}}}$

$d) \\ x {}^{2} + 4x = 1 - x + 12 \\ x {}^{2} + 4x - 13 + x = 0 \\ x {}^{2} + 5x - 13 = 0 \\ \boxed{a = 1 \:, \: b = 5 \: , \: c = - 13} \\ x = \frac{ - b ± \sqrt{b {}^{2} - 4ac } }{2a} \\ x = \frac{ - 5 ±\sqrt{5 {}^{2} - 4 \: . \: 1 \: . \: ( - 13)} }{2 \: . \: 1} \\ x = \frac{ - 5 ±\sqrt{25 - 52} }{2} \\ \boxed{\boxed{\boxed{x = \frac{ - 5± \sqrt{77} }{2} }}}$

$e) \\ x {}^{2} - x = x - 1 \\ x {}^{2} - x - x + 1 = 0 \\ x {}^{2} - 2x + 1 = 0 \\ \boxed{a = 1 \: , \: b = - 2 \:, \: c = 1} \\ x = \frac{ - b ± \sqrt{b {}^{2} - 4ac } }{2a} \\ x = \frac{ - ( - 2) ±\sqrt{( - 2) {}^{2} - 4 \: . \: 1 \: . \: 1} }{2 \: . \: 1} \\ x= \frac{ 2 ± \sqrt{4 - 4} }{2} \\ x = \frac{2 ±\sqrt{0} }{2} \\ x = \frac{2 ±0}{2} \\ x = \frac{2}{2} \\ \boxed{\boxed{\boxed{x = 1}}}$

$f) \\ 10x {}^{2} - 1 = 3x \\ 10x {}^{2} - 3x - 1 = 0 \\ \boxed{a = 10 \: ,\: b = - 3 \: , \: c = - 1} \\ x = \frac{ - b± \sqrt{b {}^{2} - 4ac} }{2a} \\ x = \frac{ - ( - 3) ±\sqrt{( - 3) {}^{2} - 4 \: . \: 10 \: . \: ( - 1)} }{2 \: . \: 10} \\ x = \frac{3 ± \sqrt{9 + 40} }{20} \\ x = \frac{3 ±\sqrt{49} }{20} \\ x = \frac{3± 7}{20} \\ x = \frac{3 + 7}{20} = \frac{10}{20} = \boxed{\boxed{\boxed{ \frac{1}{2} }}} \\ x = \frac{3 - 7}{20} = \frac{ - 4}{20} = \boxed{\boxed{\boxed{ - \frac{1}{5} }}}$

$g) \\ x {}^{2} = 3x + 4 \\ x {}^{2} - 3x - 4 = 0 \\ \boxed{a = 1 \: , \: b = - 3 \: ,\: c = - 4} \\ x = \frac{ - b ±\sqrt{b {}^{2} - 4ac} }{2a} \\ x = \frac{ - ( - 3) ± \sqrt{( - 3) {}^{2} - 4 \: . \: 1 \: . \: ( - 4)} }{2 \: . \: 1} \\ x = \frac{3 ±\sqrt{9 + 16} }{2} \\ x = \frac{3 ± \sqrt{25} }{2} \\ x = \frac{3 ± \sqrt{5} }{2} \\ x = \frac{3 + 5 }{2} = \frac{8}{2} = \boxed{\boxed{\boxed{4}}} \\ x = \frac{3 - 5}{2} = \frac{ - 2}{2} = \boxed{\boxed{\boxed{ - 1}}}$

$h) \\ 9x {}^{2} - 5x = x - 1 \\ 9x {}^{2} - 5x - x + 1 = 0 \\ 9x {}^{2} - 6x + 1 = 0 \\ \boxed{a = 9 \: , \: b = - 6 \: ,\: c = 1} \\ x = \frac{ - b± \sqrt{b {}^{2} - 4ac} }{2a} \\ x = \frac{ - ( - 6) ±\sqrt{( - 6) {}^{2} - 4 \: . \: 9 \: . \: 1} }{2 \: . \: 9} \\ x = \frac{6 ±\sqrt{36 - 36} }{18} \\ x = \frac{6 ± \sqrt{0} }{18} \\ x = \frac{6± 0}{18} \\ x = \frac{6}{18} \\ \boxed{\boxed{\boxed{x = \frac{1}{3} }}}$

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