3x^2-1x+2=0 eching

$$3 { x }^{ 2 } -1x+2=0$$

x uchun yechish (complex solution)

$$x=\frac{1+\sqrt{23}i}{6}\approx 0,166666667+0,799305254i$$

$$x=\frac{-\sqrt{23}i+1}{6}\approx 0,166666667-0,799305254i$$

Kvadrat tenglamalaridan foydalanish qoidalari

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$3x^{2}-x+2=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, -1 ni b va 2 ni c bilan almashtiring.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 3\times 2}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-\left(-1\right)±\sqrt{1-12\times 2}}{2\times 3}$$

-12 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-1\right)±\sqrt{1-24}}{2\times 3}$$

1 ni -24 ga qo'shish.

$$x=\frac{-\left(-1\right)±\sqrt{-23}}{2\times 3}$$

-23 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-1\right)±\sqrt{23}i}{2\times 3}$$

-1 ning teskarisi 1 ga teng.

$$x=\frac{1±\sqrt{23}i}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{1±\sqrt{23}i}{6}$$

x=\frac{1±\sqrt{23}i}{6} tenglamasini yeching, bunda ± musbat. 1 ni i\sqrt{23} ga qo'shish.

$$x=\frac{1+\sqrt{23}i}{6}$$

x=\frac{1±\sqrt{23}i}{6} tenglamasini yeching, bunda ± manfiy. 1 dan i\sqrt{23} ni ayirish.

$$x=\frac{-\sqrt{23}i+1}{6}$$

Tenglama yechildi.

$$x=\frac{1+\sqrt{23}i}{6} x=\frac{-\sqrt{23}i+1}{6}$$

Kvadratni yakunlash bo'yicha qadamlar

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$3x^{2}-x+2=0$$

Tenglamaning ikkala tarafidan 2 ni ayirish.

$$3x^{2}-x+2-2=-2$$

O‘zidan 2 ayirilsa 0 qoladi.

$$3x^{2}-x=-2$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}-x}{3}=-\frac{2}{3}$$

3 ga bo'lish 3 ga ko'paytirishni bekor qiladi.

$$x^{2}-\frac{1}{3}x=-\frac{2}{3}$$

-\frac{1}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{6} olish uchun. Keyin, -\frac{1}{6} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=-\frac{2}{3}+\left(-\frac{1}{6}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{6} kvadratini chiqarish.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=-\frac{2}{3}+\frac{1}{36}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{2}{3} ni \frac{1}{36} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=-\frac{23}{36}$$

x^{2}-\frac{1}{3}x+\frac{1}{36} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{6}\right)^{2}=-\frac{23}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{-\frac{23}{36}}$$

Qisqartirish.

$$x-\frac{1}{6}=\frac{\sqrt{23}i}{6} x-\frac{1}{6}=-\frac{\sqrt{23}i}{6}$$

\frac{1}{6} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{1+\sqrt{23}i}{6} x=\frac{-\sqrt{23}i+1}{6}$$