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SUEC 高中 Adv Maths (Discriminant of Roots)
- 1. 判别式 𝒑𝒈 𝟓𝟐 complex ∆ 2 equal roots 有两不等实根 无实根 有两相等实根
- 2. 判别式 𝒑𝒈 𝟓𝟐, 𝟓𝟑 𝟔. (𝒂) 𝟑𝒙𝟐 + 𝟓𝒙 − 𝟏 = 𝟎 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 𝒃 𝟓𝒙𝟐 + 𝟕 = 𝟎 𝐚 = 𝟑 𝒃 = 𝟓 𝐜 = −𝟏 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 = 𝟓𝟐 − 𝟒 𝟑 −𝟏 = 𝟐𝟓 + 𝟏𝟐 = 𝟑𝟕 > 𝟎 ∴ 𝟑𝒙𝟐 + 𝟓𝒙 − 𝟏 = 𝟎 𝒉𝒂𝒗𝒆 𝟐 𝒅𝒊𝒇𝒇. 𝒓𝒐𝒐𝒕𝒔 𝐚 = 𝟓 𝒃 = 𝟎 𝐜 = 𝟕 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 = 𝟎 − 𝟒 𝟓 𝟕 = −𝟏𝟒𝟎 < 𝟎 ∴ 𝟓𝒙𝟐 + 𝟕 = 𝟎 𝒉𝒂𝒗𝒆 𝒏𝒐 𝒓𝒆𝒂𝒍 𝒓𝒐𝒐𝒕𝒔 2 equal roots no real roots 2 different roots 无实根 有两相等实根 有两不等实根
- 3. 判别式 𝒑𝒈 𝟓𝟑 𝟔. 𝒄 𝟒𝒚𝟐 − 𝟏𝟐𝒚 + 𝟗 = 𝟎 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 𝒅 𝟑𝒛𝟐 − 𝟕𝒛 + 𝟑 = 𝟎 𝐚 = 𝟒 𝒃 = −𝟏𝟐 𝐜 = 𝟗 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 = −𝟏𝟐 𝟐 − 𝟒 𝟒 𝟗 = 𝟏𝟒𝟒 − 𝟏𝟒𝟒 = 𝟎 ∴ 𝟒𝒚𝟐 − 𝟏𝟐𝒚 + 𝟗 = 𝟎 𝒉𝒂𝒗𝒆 𝟐 𝒆𝒒𝒖𝒂𝒍 𝒓𝒐𝒐𝒕𝒔 𝐚 = 𝟑 𝒃 = −𝟕 𝐜 = 𝟑 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 = −𝟕 𝟐 − 𝟒 𝟑 𝟑 = 𝟏𝟑 > 𝟎 2 equal roots no real roots 2 different roots = 𝟒𝟗 − 𝟑𝟔 ∴ 𝟑𝒛𝟐 − 𝟕𝒛 + 𝟑 = 𝟎 𝒉𝒂𝒗𝒆 𝟐 𝒅𝒊𝒇𝒇. 𝒓𝒐𝒐𝒕𝒔 无实根 有两相等实根 有两不等实根
- 4. 判别式 𝒑𝒈 𝟓𝟒 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 2 equal roots no real roots 2 different roots 有两不等实根 有两不等实根 有两不等实根 有两不等实根 有两不等实根 有两不等实根 无实根 无实根 无实根 有两相等实根 无实根 有两相等实根 有两不等实根
- 5. 𝒂 = 𝟏 𝒃 = −(𝟐𝒌 + 𝟏) 𝐜 = (𝒌𝟐 − 𝟏) ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 = − 𝟐𝒌 − 𝟏 𝟐 −𝟒 𝟏 𝒌𝟐 − 𝟏 = 𝟒𝒌𝟐 + 𝟒𝒌 + 𝟏 − 𝟒𝒌𝟐 + 𝟒 = 𝟒𝒌 + 𝟓 𝒂 ∆ = 𝟎 𝟒𝒌 + 𝟓 = 𝟎 𝒌 = − 𝟓 𝟒 𝒃 ∆ > 𝟎 𝟒𝒌 + 𝟓 > 𝟎 𝒌 > − 𝟓 𝟒 𝒄 ∆ < 𝟎 𝟒𝒌 + 𝟓 < 𝟎 𝒌 < − 𝟓 𝟒 判别式 𝒑𝒈 𝟓𝟑 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 2 equal roots no real roots 2 different roots 7 无实根 有两相等实根 有两不等实根
- 6. 判别式 𝒑𝒈 𝟓𝟓 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 2 equal roots no real roots 2 different roots 无实根 有两相等实根 有两不等实根 𝒎 = − 𝟏 𝟒 − 𝟏 𝟒 < 𝒎 < 𝟎 𝒎 > 𝟎 𝒎 < − 𝟏 𝟒 或 𝒑 = 𝟓 𝒙𝟏 = 𝒙𝟐 = −𝟏
- 7. 𝒂 = 𝟒 𝒃 = −𝟐𝟖 𝐜 = 𝒎 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 = 𝟎 −𝟐𝟖 𝟐 − 𝟒 𝟒 𝒎 = 𝟎 7𝟖𝟒 = 𝟏𝟔𝒎 𝒎 = 𝟕𝟖𝟒 𝟏𝟔 𝒎 = 𝟒𝟗 判别式 𝒑𝒈 𝟓𝟒 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 2 equal roots no real roots 2 different roots 8 无实根 有两相等实根 有两不等实根
- 8. 判别式 𝒑𝒈 𝟓𝟓 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 2 equal roots no real roots 2 different roots 无实根 有两相等实根 有两不等实根 𝒎 = 𝟎 , 𝟓 𝒎 = 𝟎 , 𝟑
- 9. 判别式 𝒑𝒈 𝟓𝟒 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 2 equal roots no real roots 2 different roots 𝒂 = 𝒎 + 𝟑 𝒃 = − 𝒎𝟐 − 𝟖 𝐜 = 𝒎 − 𝟑 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 = − 𝒎𝟐 − 𝟖 𝟐 − 𝟒 𝒎 + 𝟑 𝒎 − 𝟑 = −𝒎𝟐 + 𝟖 𝟐 − 𝟒 𝒎𝟐 − 𝟗 = 𝒎𝟒 −𝟏𝟔𝒎𝟐 + 𝟔𝟒 − 𝟒𝒎𝟐 + 𝟑𝟔 = 𝒎𝟒 −𝟐𝟎𝒎𝟐 + 𝟏𝟎𝟎 = 𝒎𝟐 − 𝟏𝟎 𝟐 无实根 有两相等实根 有两不等实根
- 10. 𝒂 = 𝒌 + 𝟒 𝒃 = −(𝒌 + 𝟑) 𝐜 = −𝟏 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 ≥ 𝟎 − 𝒌 + 𝟑 𝟐 − 𝟒 𝒌 + 𝟒 −𝟏 ≥ 𝟎 𝒌𝟐 + 𝟗 + 𝟔𝒌 + 𝟒𝒌 + 𝟏𝟔 ≥ 𝟎 𝒌𝟐 + 𝟐𝟓 + 𝟏𝟎𝒌 ≥ 𝟎 𝒌 + 𝟓 𝟐 ≥ 𝟎 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 2 equal roots no real roots 2 different roots 无实根 有两相等实根 有两不等实根
- 11. 判别式 𝒑𝒈 𝟓𝟓 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 2 equal roots no real roots 2 different roots 无实根 有两相等实根 有两不等实根 𝒎 = 𝟔
- 12. 判别式 𝒑𝒈 𝟓𝟓 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 2 equal roots no real roots 2 different roots 无实根 有两相等实根 有两不等实根
- 13. 判别式 𝒑𝒈 𝟓𝟓 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 2 equal roots no real roots 2 different roots 无实根 有两相等实根 有两不等实根 𝒂 = 𝒑 𝒃 = 𝒑 + 𝒒 𝐜 = 𝒒 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 = 𝒑 + 𝒒 𝟐 − 𝟒 𝒑 𝒒 = 𝒑𝟐 + 𝟐𝒑𝒒 + 𝒒𝟐 − 𝟒𝒑𝒒 = 𝒑𝟐 − 𝟐𝒑𝒒 + 𝒒𝟐 = 𝒑 − 𝒒 𝟐 𝒑 − 𝒒 𝟐 ≥ 𝟎 ∴ ∆ ≥ 𝟎
- 14. 判别式 𝒑𝒈 𝟓𝟓 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 2 equal roots no real roots 2 different roots 无实根 有两相等实根 有两不等实根 𝒂 = 𝟏 − 𝒒 + 𝒑𝟐 𝟐 𝒃 = 𝒑(𝟏 + 𝒒) 𝐜 = 𝒒 𝒒 − 𝟏 + 𝒑𝟐 𝟐 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 ≥ 𝟎 ∆ = 𝒑 𝟏 + 𝒒 𝟐 − 𝟒 𝟏 − 𝒒 + 𝒑𝟐 𝟐 𝒒 𝒒 − 𝟏 + 𝒑𝟐 𝟐 ∆ 𝒘𝒊𝒍𝒍 𝒐𝒏𝒍𝒚 𝒃𝒆 𝒕𝒓𝒖𝒆 𝒊𝒇 𝒃𝟐 = 𝟒𝒂𝒄 𝑳𝒆𝒕 𝒑𝟐 = 𝟒𝒒 𝟒𝒒 𝟏 + 𝟐𝒒 + 𝒒𝟐 = 𝟒 𝟏 − 𝒒 + 𝟒𝒒 𝟐 𝒒 𝒒 − 𝟏 + 𝟒𝒒 𝟐 = 𝟒 𝟏 − 𝒒 + 𝟐𝒒 (𝒒𝟐 − 𝒒 + 𝟐𝒒) = 𝟒 𝟏 + 𝒒 (𝒒𝟐 + 𝒒) = 𝟒 𝒒𝟐 + 𝒒 + 𝒒𝟑 + 𝒒𝟐 = 𝟒𝒒 𝟏 + 𝟐𝒒 + 𝒒𝟐
- 15. 判别式 𝒑𝒈 𝟓𝟓 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 2 equal roots no real roots 2 different roots 无实根 有两相等实根 有两不等实根 𝒂𝟐 + 𝒃𝟐 − 𝒄𝟐 = 𝟎 𝒄 − 𝒂 𝟐 = 𝟒(𝒃 − 𝒄)(𝒂 − 𝒃)
- 16. 判别式 𝒑𝒈 𝟓𝟓 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 2 equal roots no real roots 2 different roots 无实根 有两相等实根 有两不等实根 𝒂𝟐 + 𝒃𝟐 − 𝒄𝟐 = 𝟎 𝒂(𝟏 − 𝒙𝟐) + 𝟐𝒃𝒙 + 𝒄 𝟏 + 𝒙𝟐 = 𝟎 𝒂 − 𝒂𝒙𝟐 + 𝟐𝒃𝒙 + 𝒄 + 𝒄𝒙𝟐 = 𝟎 (−𝒂 + 𝒄)𝒙𝟐 + 𝟐𝒃𝒙 + 𝒂 + 𝒄 = 𝟎 𝐚 = −𝒂 + 𝒄 𝒃 = 𝟐𝒃 𝐜 = 𝒂 + 𝒄 ∆ = 𝒃𝟐 − 𝟒𝒂𝒄 = 𝟎 𝟐𝒃 𝟐 − 𝟒 −𝒂 + 𝒄 𝒂 + 𝒄 = 𝟎 𝟒𝒃𝟐 − 𝟒 𝒄𝟐 − 𝒂𝟐 = 𝟎