解答
2x2−6x+6=c1(x2+x−1)+c2(x2−x+1)
解答
x=2(2−c2−c1)6−c2+c1+5c12+28c1−2c1c2+20c2−3c22−12,x=2(2−c2−c1)6−c2+c1−5c12+28c1−2c1c2+20c2−3c22−12;2−c2−c1=0
求解步骤
2x2−6x+6=c1(x2+x−1)+c2(x2−x+1)
展开 c1(x2+x−1)+c2(x2−x+1):c1x2+c1x−c1+c2x2−c2x+c2
2x2−6x+6=c1x2+c1x−c1+c2x2−c2x+c2
两边减去 −c2x+c22x2−6x+6−(−c2x+c2)=c1x2+c1x−c1+c2x2−c2x+c2−(−c2x+c2)
化简
2x2−6x+6+c2x−c2=c1x2+c1x−c1+c2x2
将 c2x2para o lado esquerdo
2x2−6x+6+c2x−c2−c2x2=c1x2+c1x−c1
两边减去 c1x−c12x2−6x+6+c2x−c2−c2x2−(c1x−c1)=c1x2+c1x−c1−(c1x−c1)
化简
2x2−6x+6+c2x−c2−c2x2−c1x+c1=c1x2
将 c1x2para o lado esquerdo
2x2−6x+6+c2x−c2−c2x2−c1x+c1−c1x2=0
改写成标准形式 ax2+bx+c=0(2−c2−c1)x2+(−6+c2−c1)x+6−c2+c1=0
使用求根公式求解
x1,2=2(2−c2−c1)−(−6+c2−c1)±(−6+c2−c1)2−4(2−c2−c1)(6−c2+c1)
化简 (−6+c2−c1)2−4(2−c2−c1)(6−c2+c1):5c12+28c1−2c1c2+20c2−3c22−12
x1,2=2(2−c2−c1)−(−6+c2−c1)±5c12+28c1−2c1c2+20c2−3c22−12;2−c2−c1=0
将解分隔开x1=2(2−c2−c1)−(−6+c2−c1)+5c12+28c1−2c1c2+20c2−3c22−12,x2=2(2−c2−c1)−(−6+c2−c1)−5c12+28c1−2c1c2+20c2−3c22−12
x=2(2−c2−c1)−(−6+c2−c1)+5c12+28c1−2c1c2+20c2−3c22−12:2(2−c2−c1)6−c2+c1+5c12+28c1−2c1c2+20c2−3c22−12
x=2(2−c2−c1)−(−6+c2−c1)−5c12+28c1−2c1c2+20c2−3c22−12:2(2−c2−c1)6−c2+c1−5c12+28c1−2c1c2+20c2−3c22−12
二次方程组的解是:x=2(2−c2−c1)6−c2+c1+5c12+28c1−2c1c2+20c2−3c22−12,x=2(2−c2−c1)6−c2+c1−5c12+28c1−2c1c2+20c2−3c22−12;2−c2−c1=0