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ax^2+bx+c=0的解為x1,2=[-b±√(b^2-4ac)]/2a若有一根為1,則[-b±√(b^2-4ac)]/2a有一根為1即√(b^2-4ac)或-√(b^2-4ac)=2a+b兩端平方得b^2-4ac=4a^2+4ab+b^2化簡得a+b+c=0由此得:ax^2+bx+c=0有一根為1的充要條…
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