Question

(x - 1)² + (2x + 1). (2x - 1) = 0
Equação de segundo grau com incógnita.​

Answer 1

Resposta:

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

Aplicando a fórmula de bhaskara

${5x}^{2} - 2x = 0 \\ \\ \frac{ - ( - 2) \frac{ + }{ - } \sqrt{ { (- 2)}^{2} - 4 \times (1) \times (0) } }{2 \times (5)} = \\ \\ \frac{ + 2 \frac{ + }{ - } \sqrt{4 - 0} }{10} = \\ \\ \frac{2 \frac{ + }{ - } \sqrt{4} }{10} = \frac{2 \frac{ + }{ - } 2}{10} \\ \\ {x}^{1} = \frac{2 + 2}{10} = \frac{4}{10} = \frac{2}{5} \\ \\ {x}^{2} = \frac{2 - 2}{10} = \frac{0}{10}$

Logo a raiz sera 4/10 e 0