3x^2+1=2x eching

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

x uchun yechish (complex solution)

$$x=\frac{1+\sqrt{2}i}{3}\approx 0,333333333+0,471404521i$$

$$x=\frac{-\sqrt{2}i+1}{3}\approx 0,333333333-0,471404521i$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan 2x ni ayirish.

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

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.

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

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

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

-2 kvadratini chiqarish.

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

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4-12}}{2\times 3}$$

4 ni -12 ga qo'shish.

$$x=\frac{-\left(-2\right)±\sqrt{-8}}{2\times 3}$$

-8 ning kvadrat ildizini chiqarish.

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

-2 ning teskarisi 2 ga teng.

$$x=\frac{2±2\sqrt{2}i}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{2±2\sqrt{2}i}{6}$$

x=\frac{2±2\sqrt{2}i}{6} tenglamasini yeching, bunda ± musbat. 2 ni 2i\sqrt{2} ga qo'shish.

$$x=\frac{2+2\sqrt{2}i}{6}$$

2+2i\sqrt{2} ni 6 ga bo'lish.

$$x=\frac{1+\sqrt{2}i}{3}$$

x=\frac{2±2\sqrt{2}i}{6} tenglamasini yeching, bunda ± manfiy. 2 dan 2i\sqrt{2} ni ayirish.

$$x=\frac{-2\sqrt{2}i+2}{6}$$

2-2i\sqrt{2} ni 6 ga bo'lish.

$$x=\frac{-\sqrt{2}i+1}{3}$$

Tenglama yechildi.

$$x=\frac{1+\sqrt{2}i}{3} x=\frac{-\sqrt{2}i+1}{3}$$

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan 2x ni ayirish.

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

Ikkala tarafdan 1 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

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

Ikki tarafini 3 ga bo‘ling.

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

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

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

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{3} kvadratini chiqarish.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=-\frac{1}{3}+\frac{1}{9}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{1}{3} ni \frac{1}{9} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=-\frac{2}{9}$$

x^{2}-\frac{2}{3}x+\frac{1}{9} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{3}\right)^{2}=-\frac{2}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{-\frac{2}{9}}$$

Qisqartirish.

$$x-\frac{1}{3}=\frac{\sqrt{2}i}{3} x-\frac{1}{3}=-\frac{\sqrt{2}i}{3}$$

\frac{1}{3} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{1+\sqrt{2}i}{3} x=\frac{-\sqrt{2}i+1}{3}$$