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Viyet teoremasi

🟢 — masalani tezlashtiradi · 🟡 — ilg'or daraja · 🔵 — chuqur o'rganish (olimpiada darajasi)

🟢 Asosiy munosabatlar

ax2+bx+c=0ax^2 + bx + c = 0 ildizlari α\alpha, β\beta bo'lsin.

α+β=−baαβ=caax2+bx+c=a(x−α)(x−β)\alpha + \beta = -\frac{b}{a} \qquad \alpha\beta = \frac{c}{a} \qquad ax^2 + bx + c = a(x - \alpha)(x - \beta)

  • Ildizlari berilgan tenglama: x2−(α+β)x+αβ=0x^2 - (\alpha + \beta)x + \alpha\beta = 0

🟢 Simmetrik ifodalar

Ifodaα+β\alpha + \beta va αβ\alpha\beta orqali
α2+β2\alpha^2 + \beta^2(α+β)2−2αβ(\alpha + \beta)^2 - 2\alpha\beta
(α−β)2(\alpha - \beta)^2(α+β)2−4αβ(\alpha + \beta)^2 - 4\alpha\beta
α3+β3\alpha^3 + \beta^3(α+β)3−3αβ(α+β)(\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)
α4+β4\alpha^4 + \beta^4(α2+β2)2−2α2β2(\alpha^2 + \beta^2)^2 - 2\alpha^2\beta^2
α2β+αβ2\alpha^2\beta + \alpha\beta^2αβ(α+β)\alpha\beta(\alpha + \beta)
1α+1β\dfrac{1}{\alpha} + \dfrac{1}{\beta}α+βαβ\dfrac{\alpha + \beta}{\alpha\beta}
αβ+βα\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha}α2+β2αβ\dfrac{\alpha^2 + \beta^2}{\alpha\beta}
  • Ildizlar ayirmasi: ∣α−β∣=D∣a∣|\alpha - \beta| = \dfrac{\sqrt{D}}{|a|}

Misol: x2−5x+6=0x^2 - 5x + 6 = 0 da α+β=5\alpha + \beta = 5, αβ=6\alpha\beta = 6, demak α3+β3=125−90=35\alpha^3 + \beta^3 = 125 - 90 = 35. Tekshirish: 8+27=358 + 27 = 35 ✓

🔵 Yangi ildizli tenglama tuzish

ax2+bx+c=0ax^2 + bx + c = 0 dan:

Yangi ildizlarYangi tenglama
−α, −β-\alpha,\ -\betaax2−bx+c=0ax^2 - bx + c = 0
1α, 1β\dfrac{1}{\alpha},\ \dfrac{1}{\beta}cx2+bx+a=0cx^2 + bx + a = 0
kα, kβk\alpha,\ k\betaax2+kbx+k2c=0ax^2 + kbx + k^2c = 0
α+h, β+h\alpha + h,\ \beta + ha(x−h)2+b(x−h)+c=0a(x - h)^2 + b(x - h) + c = 0
α2, β2\alpha^2,\ \beta^2x2−(α2+β2)x+α2β2=0x^2 - (\alpha^2 + \beta^2)x + \alpha^2\beta^2 = 0

🟢 Ildizlar ishorasi

HolatShartlar
Ikkala ildiz musbatD≥0D \ge 0,   −ba>0\;-\dfrac{b}{a} > 0,   ca>0\;\dfrac{c}{a} > 0
Ikkala ildiz manfiyD≥0D \ge 0,   −ba<0\;-\dfrac{b}{a} < 0,   ca>0\;\dfrac{c}{a} > 0
Ishoralari har xilca<0\dfrac{c}{a} < 0 (bunda D>0D > 0 avtomatik)

🔵 Kub tenglama uchun Viyet

ax3+bx2+cx+d=0ax^3 + bx^2 + cx + d = 0 ildizlari α\alpha, β\beta, γ\gamma bo'lsa:

α+β+γ=−baαβ+βγ+γα=caαβγ=−da\alpha + \beta + \gamma = -\frac{b}{a} \qquad \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \qquad \alpha\beta\gamma = -\frac{d}{a}