Question

Помогите пожалуйста с Алгеброю!
Розвяжіть рівняння!
$1) \: x {}^{2} - 3x + 2 = 0 \\ 2) \: x {}^{2} + 12x - 13 = 0 \\ 3) \: 2x {}^{2} - 7x - 4 = 0 \\ 4) \: x {}^{2} - 8x + 20 = 0$
​

Answer 1

Ответ:

1)

$x^{2} -3x + 2 = 0\\D = 9-4*2 = 9-8 = 1\\x_{1} = \frac{3-1}{2} = 1 \\\\ x_{2} = \frac{3+1}{2} = 2$

2)

$x^{2} +12x - 13 = 0\\D = 12^{2} -4*(-13) = 144+52 = 196 = 14^{2} \\x_{1} = \frac{-12+14}{2} = 1\\\\x_{2} = \frac{-12-14}{2} = -13$

3)

$2x^{2} - 7x - 4 = 0\\D= 7^{2} - 4*(-4)*2 = 49+32 = 81 = 9^{2} \\x_{1} = \frac{7+9}{2*2} = 4\\\\ x_{2} = \frac{7-9}{2*2} = -0.5$

4)

$x^{2} -8x + 20 = 0\\D = (-8)^{2} - 4*20 = 64-80 = -16\\D < 0$

значит решений нет