Re: [代數] 資優班考題1題
※ 引述《stu2005131 (自由幻夢)》之銘言：
: 解方程式[5-(5-x)^(2)]^2=x
令t=5-x =>x=5-t≧0=>5≧t
=> (5-t^2)^2=5-t
=> 25-10t^2+t^4=5-t
=> t^4-10t^2+t+20=0
=> (t^2+t-5)(t^2-t-4)=0
=> t=-1±√21/2 or t=1±√17/2
=> x=11±√21/2 or x=9±√17/2
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