2x^2+6x+8=0 eching

$$2 x ^ { 2 } + 6 x + 8 = 0$$

x uchun yechish (complex solution)

$$x=\frac{-3+\sqrt{7}i}{2}\approx -1,5+1,322875656i$$

$$x=\frac{-\sqrt{7}i-3}{2}\approx -1,5-1,322875656i$$

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.

$$2x^{2}+6x+8=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, 6 ni b va 8 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}-4\times 2\times 8}}{2\times 2}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\times 2\times 8}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-8\times 8}}{2\times 2}$$

-8 ni 8 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-64}}{2\times 2}$$

36 ni -64 ga qo'shish.

$$x=\frac{-6±\sqrt{-28}}{2\times 2}$$

-28 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±2\sqrt{7}i}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-6±2\sqrt{7}i}{4}$$

x=\frac{-6±2\sqrt{7}i}{4} tenglamasini yeching, bunda ± musbat. -6 ni 2i\sqrt{7} ga qo'shish.

$$x=\frac{-6+2\sqrt{7}i}{4}$$

-6+2i\sqrt{7} ni 4 ga bo'lish.

$$x=\frac{-3+\sqrt{7}i}{2}$$

x=\frac{-6±2\sqrt{7}i}{4} tenglamasini yeching, bunda ± manfiy. -6 dan 2i\sqrt{7} ni ayirish.

$$x=\frac{-2\sqrt{7}i-6}{4}$$

-6-2i\sqrt{7} ni 4 ga bo'lish.

$$x=\frac{-\sqrt{7}i-3}{2}$$

Tenglama yechildi.

$$x=\frac{-3+\sqrt{7}i}{2} x=\frac{-\sqrt{7}i-3}{2}$$

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.

$$2x^{2}+6x+8=0$$

Tenglamaning ikkala tarafidan 8 ni ayirish.

$$2x^{2}+6x+8-8=-8$$

O‘zidan 8 ayirilsa 0 qoladi.

$$2x^{2}+6x=-8$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+6x}{2}=-\frac{8}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{6}{2}x=-\frac{8}{2}$$

6 ni 2 ga bo'lish.

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

-8 ni 2 ga bo'lish.

$$x^{2}+3x=-4$$

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{2} kvadratini chiqarish.

$$x^{2}+3x+\frac{9}{4}=-4+\frac{9}{4}$$

-4 ni \frac{9}{4} ga qo'shish.

$$x^{2}+3x+\frac{9}{4}=-\frac{7}{4}$$

x^{2}+3x+\frac{9}{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.

$$\left(x+\frac{3}{2}\right)^{2}=-\frac{7}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{-\frac{7}{4}}$$

Qisqartirish.

$$x+\frac{3}{2}=\frac{\sqrt{7}i}{2} x+\frac{3}{2}=-\frac{\sqrt{7}i}{2}$$

Tenglamaning ikkala tarafidan \frac{3}{2} ni ayirish.

$$x=\frac{-3+\sqrt{7}i}{2} x=\frac{-\sqrt{7}i-3}{2}$$