Question

Если x1 и x2-корни уравнения 
2x^2+3x-4=0, то значение выражения x1^4+x2^4 равно ?

Answer 1

По теореме Виета 
$2x^2+3x-4=0\ x_{1}+x_{2}=-frac{3}{2}\ x_{1}x_{2}=-2\\ (x_{1}+x_{2})^4=x_{1}^4+4x_{1}x_{2}^3+6x_{1}^2x_{2}^2+4x_{1}^3x_{2}+x_{2}^4=frac{81}{16}\ 4x_{1}x_{2}(x_{2}^2+x_{1}^2)+6x_{1}^2x_{2}^2=\ x_{1}^2+x_{2}^2=(x_{1}+x_{2})^2-2x_{1}x_{2}=frac{9}{4}+2*2=frac{25}{4}\ 4x_{1}x_{2}(x_{2}^2+x_{1}^2)+6x_{1}^2x_{2}^2=4*-2(frac{25}{4}) +6*4 = -50+24 = -26\ \ x_{1}^4+x_{2}^4=frac{81}{16}+26=frac{497}{16}$