9x^2+6x+3=0 eching

$$9 { x }^{ 2 } +6x+3 = 0$$

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

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.

$$9x^{2}+6x+3=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 9 ni a, 6 ni b va 3 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}-4\times 9\times 3}}{2\times 9}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\times 9\times 3}}{2\times 9}$$

-4 ni 9 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-36\times 3}}{2\times 9}$$

-36 ni 3 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-108}}{2\times 9}$$

36 ni -108 ga qo'shish.

$$x=\frac{-6±\sqrt{-72}}{2\times 9}$$

-72 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±6\sqrt{2}i}{2\times 9}$$

2 ni 9 marotabaga ko'paytirish.

$$x=\frac{-6±6\sqrt{2}i}{18}$$

x=\frac{-6±6\sqrt{2}i}{18} tenglamasini yeching, bunda ± musbat. -6 ni 6i\sqrt{2} ga qo'shish.

$$x=\frac{-6+6\sqrt{2}i}{18}$$

-6+6i\sqrt{2} ni 18 ga bo'lish.

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

x=\frac{-6±6\sqrt{2}i}{18} tenglamasini yeching, bunda ± manfiy. -6 dan 6i\sqrt{2} ni ayirish.

$$x=\frac{-6\sqrt{2}i-6}{18}$$

-6-6i\sqrt{2} ni 18 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

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

$$9x^{2}+6x+3=0$$

Tenglamaning ikkala tarafidan 3 ni ayirish.

$$9x^{2}+6x+3-3=-3$$

O‘zidan 3 ayirilsa 0 qoladi.

$$9x^{2}+6x=-3$$

Ikki tarafini 9 ga bo‘ling.

$$\frac{9x^{2}+6x}{9}=-\frac{3}{9}$$

9 ga bo'lish 9 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{6}{9}x=-\frac{3}{9}$$

\frac{6}{9} ulushini 3 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

\frac{-3}{9} ulushini 3 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

\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.

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

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

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