해법
x2+(a+ba+aa+b)x+1=0
해법
x=2(a2+ab)−2(a2+ab)−b2+(a+b)3(a2+ab)2b2(4a5+16a4b+25a3b2+19a2b3+7ab4+b5)(a2+ab),x=2(a2+ab)−2(a2+ab)−b2−(a+b)3(a2+ab)2b2(4a5+16a4b+25a3b2+19a2b3+7ab4+b5)(a2+ab)
솔루션 단계
x2+(a+ba+aa+b)x+1=0
x2+(a+ba+aa+b)x+1 확장 :x2+a2+ab2a2x+2abx+b2x+1
x2+a2+ab2a2x+2abx+b2x+1=0
표준 양식으로 작성 ax2+bx+c=0x2+(2+a2+abb2)x+1=0
쿼드 공식으로 해결
x1,2=2⋅1−(2+a2+abb2)±(2+a2+abb2)2−4⋅1⋅1
(2+a2+abb2)2−4⋅1⋅1단순화하세요:(a+b)3(a2+ab)2b2(4a5+16a4b+25a3b2+19a2b3+7ab4+b5)
x1,2=2⋅1−(2+a2+abb2)±(a+b)3(a2+ab)2b2(4a5+16a4b+25a3b2+19a2b3+7ab4+b5)
솔루션 분리x1=2⋅1−(2+a2+abb2)+(a+b)3(a2+ab)2b2(4a5+16a4b+25a3b2+19a2b3+7ab4+b5),x2=2⋅1−(2+a2+abb2)−(a+b)3(a2+ab)2b2(4a5+16a4b+25a3b2+19a2b3+7ab4+b5)
x=2⋅1−(2+a2+abb2)+(a+b)3(a2+ab)2b2(4a5+16a4b+25a3b2+19a2b3+7ab4+b5):2(a2+ab)−2(a2+ab)−b2+(a+b)3(a2+ab)2b2(4a5+16a4b+25a3b2+19a2b3+7ab4+b5)(a2+ab)
x=2⋅1−(2+a2+abb2)−(a+b)3(a2+ab)2b2(4a5+16a4b+25a3b2+19a2b3+7ab4+b5):2(a2+ab)−2(a2+ab)−b2−(a+b)3(a2+ab)2b2(4a5+16a4b+25a3b2+19a2b3+7ab4+b5)(a2+ab)
2차 방정식의 해는 다음과 같다:x=2(a2+ab)−2(a2+ab)−b2+(a+b)3(a2+ab)2b2(4a5+16a4b+25a3b2+19a2b3+7ab4+b5)(a2+ab),x=2(a2+ab)−2(a2+ab)−b2−(a+b)3(a2+ab)2b2(4a5+16a4b+25a3b2+19a2b3+7ab4+b5)(a2+ab)