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
$m_{1}= \frac{12+6}{2}=9\\\\ m_{2}= \frac{12-6}{2}=3\\\\\\ x^{2} +3x-1=9\\\\ x^{2} +3x-10=0\\\\D=3 ^{2} - 4 * 1 * (- 10) = 9+40=49= 7^{2}\\\\ x_{1}= \frac{-3+7}{2}=2\\\\ x_{2} = \frac{-3-7}{2}=-5 \\\\\\ x^{2} +3x-1=3\\\\ x^{2} +3x-4=0\\\\D= 3^{2}-4*1*(-4)=9+16=25= 5^{2} \\\\ x_{3}= \frac{-3+5}{2}=1 \\\\ x_{4} = \frac{-3-5}{2}=-4$