해법
2x2−6x+6=c1(x2+x−1)+c2(x2−x+1)
해법
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
솔루션 단계
2x2−6x+6=c1(x2+x−1)+c2(x2−x+1)
c1(x2+x−1)+c2(x2−x+1) 확장 :c1x2+c1x−c1+c2x2−c2x+c2
2x2−6x+6=c1x2+c1x−c1+c2x2−c2x+c2
빼다 −c2x+c2 양쪽에서2x2−6x+6−(−c2x+c2)=c1x2+c1x−c1+c2x2−c2x+c2−(−c2x+c2)
단순화
2x2−6x+6+c2x−c2=c1x2+c1x−c1+c2x2
c2x2를 왼쪽으로 이동
2x2−6x+6+c2x−c2−c2x2=c1x2+c1x−c1
빼다 c1x−c1 양쪽에서2x2−6x+6+c2x−c2−c2x2−(c1x−c1)=c1x2+c1x−c1−(c1x−c1)
단순화
2x2−6x+6+c2x−c2−c2x2−c1x+c1=c1x2
c1x2를 왼쪽으로 이동
2x2−6x+6+c2x−c2−c2x2−c1x+c1−c1x2=0
표준 양식으로 작성 ax2+bx+c=0(2−c2−c1)x2+(−6+c2−c1)x+6−c2+c1=0
쿼드 공식으로 해결
x1,2=2(2−c2−c1)−(−6+c2−c1)±(−6+c2−c1)2−4(2−c2−c1)(6−c2+c1)
(−6+c2−c1)2−4(2−c2−c1)(6−c2+c1)단순화하세요: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
솔루션 분리x1=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
2차 방정식의 해는 다음과 같다: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