해법
을 위해 해결하다 t,((a−c)−(bt+dt))2=(r+t)2
해법
t=d2+b2+2db−1da−dc+ab−cb+r+d2r2+b2r2+2dbr2+2dar+2abr−2dcr−2cbr+a2+c2−2ac,t=d2+b2+2db−1da−dc+ab−cb+r−d2r2+b2r2+2dbr2+2dar+2abr−2dcr−2cbr+a2+c2−2ac;d2+b2+2db−1=0
솔루션 단계
((a−c)−(bt+dt))2=(r+t)2
((a−c)−(bt+dt))2 확장 :d2t2−2dta+2dtc−2tab+2tcb+t2b2+2dt2b+a2−2ac+c2
(r+t)2 확장 :r2+2tr+t2
d2t2−2dta+2dtc−2tab+2tcb+t2b2+2dt2b+a2−2ac+c2=r2+2tr+t2
t2를 왼쪽으로 이동
d2t2−2dta+2dtc−2tab+2tcb+t2b2+2dt2b+a2−2ac+c2−t2=r2+2tr
2tr를 왼쪽으로 이동
d2t2−2dta+2dtc−2tab+2tcb+t2b2+2dt2b+a2−2ac+c2−t2−2tr=r2
r2를 왼쪽으로 이동
d2t2−2dta+2dtc−2tab+2tcb+t2b2+2dt2b+a2−2ac+c2−t2−2tr−r2=0
표준 양식으로 작성 ax2+bx+c=0(d2+b2+2db−1)t2+(−2da+2dc−2ab+2cb−2r)t+a2−2ac+c2−r2=0
쿼드 공식으로 해결
t1,2=2(d2+b2+2db−1)−(−2da+2dc−2ab+2cb−2r)±(−2da+2dc−2ab+2cb−2r)2−4(d2+b2+2db−1)(a2−2ac+c2−r2)
(−2da+2dc−2ab+2cb−2r)2−4(d2+b2+2db−1)(a2−2ac+c2−r2)단순화하세요:2a2−2ac+2dar+2abr+c2+d2r2+b2r2+2dbr2−2dcr−2cbr
t1,2=2(d2+b2+2db−1)−(−2da+2dc−2ab+2cb−2r)±2a2−2ac+2dar+2abr+c2+d2r2+b2r2+2dbr2−2dcr−2cbr;d2+b2+2db−1=0
솔루션 분리t1=2(d2+b2+2db−1)−(−2da+2dc−2ab+2cb−2r)+2a2−2ac+2dar+2abr+c2+d2r2+b2r2+2dbr2−2dcr−2cbr,t2=2(d2+b2+2db−1)−(−2da+2dc−2ab+2cb−2r)−2a2−2ac+2dar+2abr+c2+d2r2+b2r2+2dbr2−2dcr−2cbr
t=2(d2+b2+2db−1)−(−2da+2dc−2ab+2cb−2r)+2a2−2ac+2dar+2abr+c2+d2r2+b2r2+2dbr2−2dcr−2cbr:d2+b2+2db−1da−dc+ab−cb+r+d2r2+b2r2+2dbr2+2dar+2abr−2dcr−2cbr+a2+c2−2ac
t=2(d2+b2+2db−1)−(−2da+2dc−2ab+2cb−2r)−2a2−2ac+2dar+2abr+c2+d2r2+b2r2+2dbr2−2dcr−2cbr:d2+b2+2db−1da−dc+ab−cb+r−d2r2+b2r2+2dbr2+2dar+2abr−2dcr−2cbr+a2+c2−2ac
2차 방정식의 해는 다음과 같다:t=d2+b2+2db−1da−dc+ab−cb+r+d2r2+b2r2+2dbr2+2dar+2abr−2dcr−2cbr+a2+c2−2ac,t=d2+b2+2db−1da−dc+ab−cb+r−d2r2+b2r2+2dbr2+2dar+2abr−2dcr−2cbr+a2+c2−2ac;d2+b2+2db−1=0