Question

Решите прошу, с решением ​

Answer 1

Ответ:

${x}^{2} + x = 0 \\ x(x + 1) = 0 \\ x1 = 0 \\ x2 = - 1$

${x}^{2} - 4x + 3 = 0 \\ d = 16 - 4 \times 3 = 4 \\ x1 = \frac{4 + 2}{2} = 3 \\ x2 = \frac{4 - 2}{2} = 1$

${5x}^{2} + 14x - 3 = 0 \\ d = 196 - 4 \times 5 \times ( - 3) = 256 \\ x1 = \frac{ - 14 + 16}{10} = 0.2 \\ x2 = \frac{ - 14 - 16}{10 } = 3$

${x}^{2} - 2x - 2 = 0 \\d = 4 - 4 \times ( - 2) = 12 \\ x1 = \frac{2 + \sqrt{12} }{2} = \frac{2(1 + \sqrt{3)} }{2} = 1 + \sqrt{3} \\ x2 = \frac{ 2 - \sqrt{12} }{2} = \frac{2( 1 - \sqrt{3} )}{2} = 1 - \sqrt{3}$

$5x = {3x}^{2} \\3 {x}^{2} - 5x = 0 \\x(3x - 5) = 0 \\ x 1= 0 \\ x2 = \frac{5}{3}$

${x}^{2} - 5x + 4 = 0 \\ d = 25 - 4 \times 4 = 9 \\ x1 = \frac{5 + 3}{2} = 4 \\ x2 = \frac{5 - 3}{2} = 1$

${7x}^{2} - 4 = 0 \\ \sqrt{7} x = 2 \\ \sqrt{ - 7} x = - 2 \\ x1 = \frac{2}{ \sqrt{7} } \\ x2 = \frac{ - 2}{ \sqrt{ - 7} }$

$3 {x}^{2} - x + 2 = 0 \\ d = 1 - 4 \times 3 \times 2 = - 23 < 0$

Нет решения в последнем уравнении