Question

Знайдіть суму коренів рівняння: x³-3x²-4x+12=0​

Answer 1

Первый способ :

$\displaystyle\bf\\x^{3} -3x^{2} -4x+12=0\\\\(x^{3} -3x^{2}) -(4x-12)=0\\\\x^{2} \cdot(x-3)-4\cdot(x-3)=0\\\\(x-3)\cdot(x^{2} -4)=0\\\\(x-3)\cdot(x-2)\cdot(x+2)=0\\\\\\\left[\begin{array}{ccc}x-3=0\\x-2=0\\x+2=0\end{array}\right\\\\\\\left[\begin{array}{ccc}x_{1} =3\\x_{2} =2\\x_{3} =-2\end{array}\right\\\\\\x_{1} +x_{2} + x_{3} =3+2+(-2)=3$

Второй способ :

$\displaystyle\bf\\x^{3} -3x^{2} -4x+12=0\\\\a=1 \ ; \ b=-3 \ ; \ c=-4 \ ; \ d=12$

По теореме Виета для кубического уравнения :

$\displaystyle\bf\\x_{1}+ x_{2} + x_{3}=-\frac{b}{a} =-\frac{-3}{1} =3\\\\Otvet \ : \ x_{1}+ x_{2} + x_{3}=3$