[질문] x제곱-2ax+22 판별식으로계산좀요

짝수판별식d/4=(b')^2-ac=(a)^2-1*22=a^2-22이므로 a^2>22이면서로다른두실근,a^2=22이면중근,a^2<22실근이없거나허근이존재합니다(복소수의경우)