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

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

x uchun yechish (complex solution)

$$x=\frac{-1+2\sqrt{2}i}{3}\approx -0,333333333+0,942809042i$$

$$x=\frac{-2\sqrt{2}i-1}{3}\approx -0,333333333-0,942809042i$$

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.

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

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 9 ni c bilan almashtiring.

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

6 kvadratini chiqarish.

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

-4 ni 9 marotabaga ko'paytirish.

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

-36 ni 9 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-324}}{2\times 9}$$

36 ni -324 ga qo'shish.

$$x=\frac{-6±\sqrt{-288}}{2\times 9}$$

-288 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±12\sqrt{2}i}{2\times 9}$$

2 ni 9 marotabaga ko'paytirish.

$$x=\frac{-6±12\sqrt{2}i}{18}$$

x=\frac{-6±12\sqrt{2}i}{18} tenglamasini yeching, bunda ± musbat. -6 ni 12i\sqrt{2} ga qo'shish.

$$x=\frac{-6+12\sqrt{2}i}{18}$$

-6+12i\sqrt{2} ni 18 ga bo'lish.

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

x=\frac{-6±12\sqrt{2}i}{18} tenglamasini yeching, bunda ± manfiy. -6 dan 12i\sqrt{2} ni ayirish.

$$x=\frac{-12\sqrt{2}i-6}{18}$$

-6-12i\sqrt{2} ni 18 ga bo'lish.

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

Tenglama yechildi.

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

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.

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

Tenglamaning ikkala tarafidan 9 ni ayirish.

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

O‘zidan 9 ayirilsa 0 qoladi.

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

Ikki tarafini 9 ga bo‘ling.

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

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

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

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

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

-9 ni 9 ga bo'lish.

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

\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.

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

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

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

-1 ni \frac{1}{9} ga qo'shish.

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

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.

$$\left(x+\frac{1}{3}\right)^{2}=-\frac{8}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{-\frac{8}{9}}$$

Qisqartirish.

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

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

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