Diket x1dan x2 adalah akar akar persamaan kuadrat x²+6x+2=0.nilai x1²+x2²-4x1x2adalah

[tex] {x}^{2} + 6x + 2 = 0 \\ a =1 \: \: \: b = 6 \: \: \: c = 2 \\ x1 + x2 = - \frac{b}{a} = - \frac{6}{1} = - 6 \\ x1 \times x2 = \frac{c}{a} = \frac{2}{1} = 2 \\ {x1}^{2} + {x2}^{2} - 4x1x2 \\ = ({(x1 + x2)}^{2} - 2x1x2) - 4x1x2 \\ = {(x1 + x2)}^{2} - 6x1x2 \\ = {( - 6)}^{2} - 6(2) \\ = 36 - 12 \\ = 24[/tex]