Question

Уравнение и его корни

Answer 1

x²+3x=a
a²-52a-108=0
a1+a2=52 U a1*a2=-108
a1=54⇒x²+3x=54
x²+3x-54=0
x1+x2=-3 U x1*x2=-54⇒x1=-9 U x2=6
a2=-2⇒x²+3x=-2
x²+3x+2=0
x3+x4=-3 U x3*x4=2
x3=-2 U x2=-1

Answer 2

$t =(x^2 +3x) \\ \\ t^2 -52t-108=0 \\ \\ t_{1,2}=\frac{52 \pm \sqrt{52^2 - 4 \cdot 1 \cdot (-108)}}{2 \cdot 1}=\frac{52 \pm \sqrt{2704 +432}}{2}=\frac{52 \pm 56}{2}; \ \ \ t_1=54, \ t_2 =-2 \\ \\ x^2 +3x=54; \ \ \ \ x^2+3x=-2 \\ \\ x^2+3x-54=0; \ \ \ \ x^2+3x+2=0 \\ \\ x_{1,2} = \frac{-3 \pm \sqrt{9+216}}{2}; \ \ \ \ \ \ \ x_{3,4} = \frac{-3 \pm \sqrt{9-8}}{2} \\ \\ x_1=6; \ \ x_2 =-9; \ \ \ \ x_3=-1; \ \ x_4= -2$