Question

$(x {}^{2} + 3x - 1) {}^{2} - 12 {x}^{2} - 36x + 39 = 0$

Answer 1

$( x^{2} +3x-1) ^{2} -12 x^{2} -36x+39=0\\( x^{2} +3x-1) ^{2} -12( x^{2} +3x)+39=0$
Обозначим x² + 3x - 1 = m , тогда x² + 3x = m + 1
m² - 12(m + 1) + 39 = 0
m² - 12m - 12 + 39 = 0
m² - 12m + 27 = 0
D = (-12)² - 4 * 1 * 27 = 144 - 108 = 36 = 6²
$m_{1}= frac{12+6}{2}=9\\ m_{2}= frac{12-6}{2}=3\\\ x^{2} +3x-1=9\\ x^{2} +3x-10=0\\D=3 ^{2} - 4 * 1 * (- 10) = 9+40=49= 7^{2}\\ x_{1}= frac{-3+7}{2}=2\\ x_{2} = frac{-3-7}{2}=-5 \\\ x^{2} +3x-1=3\\ x^{2} +3x-4=0\\D= 3^{2}-4*1*(-4)=9+16=25= 5^{2} \\ x_{3}= frac{-3+5}{2}=1 \\ x_{4} = frac{-3-5}{2}=-4$