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SUEC 高中 Adv Maths (2 Roots Part 1)
- 1. 𝒑𝒈 𝟓𝟓 𝑥 − 𝛼 𝑥 − 𝛽 = 0 𝑥2 − ∝ +𝛽 𝑥+ ∝ 𝛽 = 0 𝑥2 − 𝛽𝑥 − 𝛼𝑥+ ∝ 𝛽 = 0
- 2. 𝒑𝒈 𝟓𝟖 𝜶 + 𝜷 = 𝟓 , 𝜶𝜷 = 𝟐 𝒂 𝒙𝟐 − 𝟓𝒙 + 𝟐 = 𝟎 𝒃 𝒙𝟐 = 𝟐𝒙 + 𝟏 𝒄 𝟑𝒙𝟐 − 𝟐𝒙 − 𝟏 = 𝟎 𝒅 𝟐𝒙𝟐 − 𝟕𝒙 = 𝟎 𝒆 𝟓𝒙𝟐 = 𝟒 𝒇 𝒙 − 𝟏 𝒙 + 𝟐 = 𝟑𝒙 𝜶 + 𝜷 = 𝟐 , 𝜶𝜷 = −𝟏 𝜶 + 𝜷 = 𝟐 𝟑 , 𝜶𝜷 = − 𝟏 𝟑 𝜶 + 𝜷 = 𝟕 𝟐 , 𝜶𝜷 = 𝟎 𝜶 + 𝜷 = 𝟎 , 𝜶𝜷 = − 𝟒 𝟓 𝜶 + 𝜷 = 𝟐 , 𝜶𝜷 = −𝟐
- 3. 𝒑𝒈 𝟓𝟕 𝒂𝒔𝒔𝒖𝒎𝒆 ∝ = −𝟏 𝒌 = ? 𝒂 = 𝟑 𝒃 = 𝟏𝟎 𝐜 = 𝒌 ∝ + 𝜷 = − 𝒃 𝒂 −𝟏 + 𝜷 = − 𝟏𝟎 𝟑 𝜷 = − 𝟕 𝟑 ∝ 𝜷 = 𝒄 𝒂 𝟕 𝟑 = 𝒌 𝟑 𝒌 = 𝟕
- 4. 𝒑𝒈 𝟓𝟕 𝒂𝒔𝒔𝒖𝒎𝒆 ∝ = −𝟒 𝛃, 𝒂 = ? 𝒂 = 𝒂 𝒃 = 𝟗 𝐜 = 𝟐𝒂 ∝ + 𝜷 = − 𝒃 𝒂 −𝟒 − 𝟏 𝟐 = − 𝟗 𝒂 − 𝟗 𝟐 = − 𝟗 𝒂 ∝ 𝜷 = 𝒄 𝒂 − 𝟒𝜷 = 𝟐𝒂 𝒂 𝜷 = − 𝟏 𝟐 𝒂 = 𝟐
- 5. 𝒑𝒈 𝟓𝟖 𝜷 = 𝟐𝟎 , 𝒎 = −𝟐𝟎 𝜷 = 𝟐 , 𝒏 = 𝟐 𝜷 = −𝟓 , 𝒎 = 𝟏𝟒 𝜷 = −𝟔 , 𝒂 = − 𝟏 𝟔 𝜷 = −𝟏 , 𝒂 = 𝟒
- 6. 𝒑𝒈 𝟓𝟖 𝒑 = 𝟎 , 𝒂 = 𝟏 𝟐 𝒒 = 𝟏 , 𝒂 = 𝟏
- 7. 𝒑𝒈 𝟓𝟖 𝒂𝒔𝒔𝒖𝒎𝒆 𝜷 = 𝜶 + 𝟐 𝒎 = ? 𝒂 = 𝟏 𝒃 = −𝟔 𝐜 = 𝒎 ∝ + 𝜷 = − 𝒃 𝒂 ∝ + ∝ +𝟐 = 𝟔 𝟐 ∝ = 𝟔 − 𝟐 ∝ = 𝟐 ∝ 𝜷 = 𝒄 𝒂 𝟐 𝟒 = 𝒎 𝒎 = 𝟖 13 𝜷 = 𝟒
- 8. 𝒑𝒈 𝟓𝟖 𝟒𝒎𝟐 = 𝟒𝒏 + 𝟏 𝒙 = 𝟑 𝟒 或 𝟑 𝟐
- 9. 𝒑𝒈 𝟓𝟖 𝒂𝒔𝒔𝒖𝒎𝒆 𝜷 = 𝜶 + 𝟏 𝒎, 𝒏 𝒆𝒒 = ? ∝ + 𝜷 = − 𝒃 𝒂 ∝ + ∝ +𝟏 = −𝟐𝒎 𝟐 ∝ = −𝟐𝒎 − 𝟏 ∝ = −𝟐𝒎 − 𝟏 𝟐 ∝ 𝜷 = 𝒄 𝒂 −𝒎 − 𝟏 𝟐 −𝒎 + 𝟏 𝟐 = 𝒏 𝜷 = −𝒎 + 𝟏 𝟐 𝒂 = 𝟏 𝒃 = 𝟐𝒎 𝐜 = 𝒏 ∝ = −𝒎 − 𝟏 𝟐 𝒎𝟐 − 𝟏 𝟒 = 𝒏 𝟒𝒎𝟐 = 𝟒𝒏 + 𝟏 𝟒𝒎𝟐 = 𝟒𝒏 + 𝟏