Question

resolva as equações de segundo grau:

a) (x+5)² + (2x-1)² = 26


b) (x-1)² + (2x+1) (2x-1) = 0


c) x (9x-8) = (x-4)​

Answer 1

a)

${(x + 5)}^{2} + {(2x - 1)}^{2} = 26 \\ \\ {x}^{2} + 10x + 25 + {4x}^{2} - 4x + 1 = 26 \\ \\ {5x}^{2} + 6x + 26 = 26 \\ \\ {5x}^{2} + 6x = 0 \\ \\ \\ \\ x = \frac{ - 6± \sqrt{36} }{10} \\ \\ x = \frac{ - 6±6}{10 } \\ \\ x' = \frac{ - 6 + 6}{10} = \frac{0}{10} = 0 \\ \\ \\ x'' = \frac{ - 6 - 6}{10} = \frac{ - 12}{10} = - \frac{6}{5}$

b)

$(x - 1) {}^{2} + (2x + 1)(2x - 1) = 0 \\ \\ {x}^{2} - 2x + 1 + {4x}^{2} - 1 = 0 \\ \\ 5x {}^{2} - 2x = 0 \\ \\ x(5x - 2) = 0 \\ \\ x' = 0 \\ \\5x - 2 = 0 \\ \\ 5x = 2 \\ \\ x = \frac{2}{5} \\ \\ x'' = \frac{2}{5}$

c)

$x(9x - 8) = (x - 4) \\ \\ 9x {}^{2} - 8x = x - 4 \\ \\ 9x {}^{2} - 8x - x + 4 = 0 \\ \\ 9 {x}^{2} - 9x + 4 = 0 \\ \\ \\ x = \frac{9± \sqrt{45} }{18} \\ \\ x = \frac{9±3 \sqrt{5} }{18} \\ \\ x = \frac{1}{2} ± \frac{3 \sqrt{5} }{18} \\ \\ x = \frac{1}{2} ± \frac{1}{6} \sqrt{5} \\ \\ x' = \frac{1}{2} + \frac{ \sqrt{5} }{6} \\ \\ x'' = \frac{1}{2} - \frac{ \sqrt{5} }{6}$