Question

При каком значении k сумма кубов корней уравнения kx^2-6kx+2k+3=0 равна 72

Answer 1

$x_{1};x_{2}$-корни уравнения. 
 $x_{1}^3+x_{2}^3=(x_{1}+x_{2})(x_{1}^2-x_{1}x_{2}+x_{2}^2)\ x_{1}^2+x_{2}^2=(x_{1}+x_{2})^2-2x_{1}x_{2}\\ x_{1}^3+x_{2}^3=(x_{1}+x_{2})((x_{1}+x_{2})^2-3x_{1}x_{2})\\ x_{1}*x_{2}=frac{2k+3}{k}\ x_{1}+x_{2}=6\\ 6(36-frac{6k+9}{k})=72\ k=frac{1}{2}$