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

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

x uchun yechish (complex solution)

$$x=\frac{-1+\sqrt{11}i}{2}\approx -0,5+1,658312395i$$

$$x=\frac{-\sqrt{11}i-1}{2}\approx -0,5-1,658312395i$$

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.

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

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

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

3 kvadratini chiqarish.

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

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9-12\times 9}}{2\times 3}$$

-12 ni 9 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9-108}}{2\times 3}$$

9 ni -108 ga qo'shish.

$$x=\frac{-3±\sqrt{-99}}{2\times 3}$$

-99 ning kvadrat ildizini chiqarish.

$$x=\frac{-3±3\sqrt{11}i}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-3±3\sqrt{11}i}{6}$$

x=\frac{-3±3\sqrt{11}i}{6} tenglamasini yeching, bunda ± musbat. -3 ni 3i\sqrt{11} ga qo'shish.

$$x=\frac{-3+3\sqrt{11}i}{6}$$

-3+3i\sqrt{11} ni 6 ga bo'lish.

$$x=\frac{-1+\sqrt{11}i}{2}$$

x=\frac{-3±3\sqrt{11}i}{6} tenglamasini yeching, bunda ± manfiy. -3 dan 3i\sqrt{11} ni ayirish.

$$x=\frac{-3\sqrt{11}i-3}{6}$$

-3-3i\sqrt{11} ni 6 ga bo'lish.

$$x=\frac{-\sqrt{11}i-1}{2}$$

Tenglama yechildi.

$$x=\frac{-1+\sqrt{11}i}{2} x=\frac{-\sqrt{11}i-1}{2}$$

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.

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

Tenglamaning ikkala tarafidan 9 ni ayirish.

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

O‘zidan 9 ayirilsa 0 qoladi.

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

Ikki tarafini 3 ga bo‘ling.

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

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

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

3 ni 3 ga bo'lish.

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

-9 ni 3 ga bo'lish.

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

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{2} kvadratini chiqarish.

$$x^{2}+x+\frac{1}{4}=-3+\frac{1}{4}$$

-3 ni \frac{1}{4} ga qo'shish.

$$x^{2}+x+\frac{1}{4}=-\frac{11}{4}$$

x^{2}+x+\frac{1}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{1}{2}\right)^{2}=-\frac{11}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{-\frac{11}{4}}$$

Qisqartirish.

$$x+\frac{1}{2}=\frac{\sqrt{11}i}{2} x+\frac{1}{2}=-\frac{\sqrt{11}i}{2}$$

Tenglamaning ikkala tarafidan \frac{1}{2} ni ayirish.

$$x=\frac{-1+\sqrt{11}i}{2} x=\frac{-\sqrt{11}i-1}{2}$$