Question

Encontre as raízes das equações do 2° grau.

$a) {x}^{2} - 6x +8 =0$

$b) {x}^{2} - 8x + 16 = 0$

$c) {x}^{2} -7x + 10 = 0$

$d) {x}^{2} + 5x - 36 = 0$

$e) {2x}^{2} + 3x - 5 = 0$


$f) {4x}^{2} - 5x - 6 = 0$

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Answer 1

Resposta:

a)$x_{1} = 2$; $x_{2} = 4$

Explicação passo a passo:

Primeiro devemos saber a fórmula de bhaskara:

$\frac{-b-+\sqrt{b^{2}-4ac} }{2a}$

Sendo:

$ax^{2} + bx +c$

a) a = 1 b = - 6 c = 8

$\frac{-(-6)-+\sqrt{6^{2}-4.1.8} }{2.1}\\\\\frac{6-+\sqrt{36-32} }{2} = \frac{6-+\sqrt{4} }{2}\\\\\frac{6+2}{2} = 4\\\\\frac{6-2}{2} = 2$

b) a = 1 b = - 8 c = 16

$\frac{-(-8)-+\sqrt{(-8)^{2}-4.1.16} }{2.1}\\\\\frac{8-+\sqrt{64-64} }{2} = \frac{8-+0}{2 } \\ \\\frac{8}{2} = 4$

c) a = 1 b = - 7 c = 10

$\frac{-(-7)-+\sqrt{7^{2}-4.1.10} }{2.1}\\\\\frac{7-+\sqrt{49-40} }{2} = \frac{7-+\sqrt{9} }{2}\\\\\frac{7+3}{2} = 5\\\\\frac{7-3}{2} = 2$

d) a = 1 b = 5 c = -36

$\frac{-(5)-+\sqrt{5^{2}-4.1.-36} }{2.1}\\\\\frac{-5-+\sqrt{25+144} }{2} = \frac{-5-+\sqrt{169} }{2}\\\\\frac{-5+13}{2} = 4\\\\\frac{-5-13}{x} = -9$

e) a = 2 b) = 3 c) = - 5

$\frac{-(3)-+\sqrt{3^{2}-4.2.-5} }{2.2}\\\\\frac{-3-+\sqrt{9 + 40} }{4} = \frac{-3-+\sqrt{49} }{4}\\\\\frac{-3+7}{4} = 1\\\\\frac{-3-7}{4} = -2,5$

f) a = 4 b) = - 5 c = - 6

$\frac{-(-5)-+\sqrt{5^{2}-4.4.-6} }{2.4}\\\\\frac{5-+\sqrt{25+96} }{8} = \frac{5-+\sqrt{121} }{8}\\\\\frac{5+11}{8} = 2\\\\\frac{5-11}{8} = - 0,75$

Espero ter ajudado~~