先由韋達定理易證 |b|<4, 考慮x在區間(-2,2)內, 令f(x)=-ax-b=x^2<4,
當a>=0時, max(f(x))=2a-b<4, 當a<0時, max(f(x))=-2a-b<4, 總之 2|a|-b<4, 即2|a|<4+b
先由韋達定理易證 |b|<4, 考慮x在區間(-2,2)內, 令f(x)=-ax-b=x^2<4,
當a>=0時, max(f(x))=2a-b<4, 當a<0時, max(f(x))=-2a-b<4, 總之 2|a|-b<4, 即2|a|<4+b
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