3x^2+2x+8=0 eching

$$3 x ^ { 2 } + 2 x + 8 = 0$$

x uchun yechish (complex solution)

$$x=\frac{-1+\sqrt{23}i}{3}\approx -0,333333333+1,598610508i$$

$$x=\frac{-\sqrt{23}i-1}{3}\approx -0,333333333-1,598610508i$$

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.

$$3x^{2}+2x+8=0$$

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 8 ni c bilan almashtiring.

$$x=\frac{-2±\sqrt{2^{2}-4\times 3\times 8}}{2\times 3}$$

2 kvadratini chiqarish.

$$x=\frac{-2±\sqrt{4-4\times 3\times 8}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4-12\times 8}}{2\times 3}$$

-12 ni 8 marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4-96}}{2\times 3}$$

4 ni -96 ga qo'shish.

$$x=\frac{-2±\sqrt{-92}}{2\times 3}$$

-92 ning kvadrat ildizini chiqarish.

$$x=\frac{-2±2\sqrt{23}i}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-2±2\sqrt{23}i}{6}$$

x=\frac{-2±2\sqrt{23}i}{6} tenglamasini yeching, bunda ± musbat. -2 ni 2i\sqrt{23} ga qo'shish.

$$x=\frac{-2+2\sqrt{23}i}{6}$$

-2+2i\sqrt{23} ni 6 ga bo'lish.

$$x=\frac{-1+\sqrt{23}i}{3}$$

x=\frac{-2±2\sqrt{23}i}{6} tenglamasini yeching, bunda ± manfiy. -2 dan 2i\sqrt{23} ni ayirish.

$$x=\frac{-2\sqrt{23}i-2}{6}$$

-2-2i\sqrt{23} ni 6 ga bo'lish.

$$x=\frac{-\sqrt{23}i-1}{3}$$

Tenglama yechildi.

$$x=\frac{-1+\sqrt{23}i}{3} x=\frac{-\sqrt{23}i-1}{3}$$

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.

$$3x^{2}+2x+8=0$$

Tenglamaning ikkala tarafidan 8 ni ayirish.

$$3x^{2}+2x+8-8=-8$$

O‘zidan 8 ayirilsa 0 qoladi.

$$3x^{2}+2x=-8$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}+2x}{3}=-\frac{8}{3}$$

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

$$x^{2}+\frac{2}{3}x=-\frac{8}{3}$$

\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.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=-\frac{8}{3}+\left(\frac{1}{3}\right)^{2}$$

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

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=-\frac{8}{3}+\frac{1}{9}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{8}{3} ni \frac{1}{9} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=-\frac{23}{9}$$

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.

$$\left(x+\frac{1}{3}\right)^{2}=-\frac{23}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{-\frac{23}{9}}$$

Qisqartirish.

$$x+\frac{1}{3}=\frac{\sqrt{23}i}{3} x+\frac{1}{3}=-\frac{\sqrt{23}i}{3}$$

Tenglamaning ikkala tarafidan \frac{1}{3} ni ayirish.

$$x=\frac{-1+\sqrt{23}i}{3} x=\frac{-\sqrt{23}i-1}{3}$$