Question

ПОМОГИТЕ ПО АЛГЕБРЕ, ПОЖАЛУЙСТА ​

Answer 1

a)

${x}^{2} + 6x = 0 \\ x(x + 6) = 0 \\ x = 0 \\ x = - 6$

б)

$- 3 {x}^{2} = 18x \\ 3x {}^{2} + 18x = 0 \\ 3x(x + 6) = 0 \\ x = 0 \\ x = - 6$

в)

$24 - 6 {x}^{2} = 0 \\ 4 - {x}^{2} = 0 \\ x {}^{2} = 4 \\ x = + - 2$

$5 {x}^{2} - 30 = 0 \\ x {}^{2} = 6 \\ x = + - \sqrt{6}$

$6 {x}^{2} - 13x - 15 = 0 \\ d = \sqrt{169 + 360} = 23 \\ x1 = \frac{13 + 23}{12} = \frac{36}{12} =2\frac{1}{6} \\ x2 = \frac{13 - 23}{12} = \frac{ - 10}{12} = - \frac{ 5}{6}$

${x}^{4} - 2 {x}^{2} - 8 = 0$

Замена

${x}^{2} = t$

$t {}^{2} - 2t - 8 = 0$

По теорема Виета корни уравнения равны 4 и - 2