Question

Решите эти два уравнения по методу выделения полного квадрата.​

Answer 1

Ответ:

1).

$x_{1} = - 6 \\ x_{2} = 2$

2).

$x_{1} = - 5 \\ x_{2} = 3$

Объяснение:

${x}^{2} + 4x - 12 = 0 \\ {x}^{2} + 2 \times x \times 2 + {2}^{2} - {2}^{2} - 12 = 0 \\ ( {x}^{2} + 2 \times x \times 2 + {2}^{2}) - 4 - 12 = 0 \\ {(x + 2)}^{2} - 16 = 0 \\ {(x + 2)}^{2} - {4}^{2} = 0 \\ ((x + 2) + 4) \times ((x + 2) - 4) = 0 \\ (x + 6) \times (x - 2) = 0 \\ x + 6 = 0 \: ili \: x - 2 = 0 \\ x_{1} = - 6 \\ x_{2} = 2$

${x}^{2} + 2x - 15 = 0 \\ {x}^{2} + 2 \times x \times 1 + {1}^{2} - {1}^{2} - 15 = 0 \\ ( {x}^{2} + 2 \times x \times 1 + {1}^{2}) - 1 - 15 = 0 \\ {(x + 1)}^{2} - 16 = 0 \\ {(x + 1)}^{2} - {4}^{2} = 0 \\ ((x + 1) + 4) \times ((x + 1) - 4) = 0 \\ (x + 5) \times (x - 3) = 0 \\ x + 5 = 0 \: \: ili \: \: x - 3 = 0 \\ x_{1} = - 5 \\ x_{2} = 3$