5x^2+6x+2=0 eching

$$5 x ^ { 2 } + 6 x + 2 = 0$$

x uchun yechish (complex solution)

$$x=-\frac{3}{5}+\frac{1}{5}i=-0,6+0,2i$$

$$x=-\frac{3}{5}-\frac{1}{5}i=-0,6-0,2i$$

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.

$$5x^{2}+6x+2=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 6 ni b va 2 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}-4\times 5\times 2}}{2\times 5}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\times 5\times 2}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-20\times 2}}{2\times 5}$$

-20 ni 2 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-40}}{2\times 5}$$

36 ni -40 ga qo'shish.

$$x=\frac{-6±\sqrt{-4}}{2\times 5}$$

-4 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±2i}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-6±2i}{10}$$

x=\frac{-6±2i}{10} tenglamasini yeching, bunda ± musbat. -6 ni 2i ga qo'shish.

$$x=\frac{-6+2i}{10}$$

-6+2i ni 10 ga bo'lish.

$$x=-\frac{3}{5}+\frac{1}{5}i$$

x=\frac{-6±2i}{10} tenglamasini yeching, bunda ± manfiy. -6 dan 2i ni ayirish.

$$x=\frac{-6-2i}{10}$$

-6-2i ni 10 ga bo'lish.

$$x=-\frac{3}{5}-\frac{1}{5}i$$

Tenglama yechildi.

$$x=-\frac{3}{5}+\frac{1}{5}i x=-\frac{3}{5}-\frac{1}{5}i$$

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.

$$5x^{2}+6x+2=0$$

Tenglamaning ikkala tarafidan 2 ni ayirish.

$$5x^{2}+6x+2-2=-2$$

O‘zidan 2 ayirilsa 0 qoladi.

$$5x^{2}+6x=-2$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}+6x}{5}=-\frac{2}{5}$$

5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{6}{5}x=-\frac{2}{5}$$

\frac{6}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{3}{5} olish uchun. Keyin, \frac{3}{5} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{6}{5}x+\left(\frac{3}{5}\right)^{2}=-\frac{2}{5}+\left(\frac{3}{5}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{5} kvadratini chiqarish.

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=-\frac{2}{5}+\frac{9}{25}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{2}{5} ni \frac{9}{25} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=-\frac{1}{25}$$

x^{2}+\frac{6}{5}x+\frac{9}{25} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{3}{5}\right)^{2}=-\frac{1}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{5}\right)^{2}}=\sqrt{-\frac{1}{25}}$$

Qisqartirish.

$$x+\frac{3}{5}=\frac{1}{5}i x+\frac{3}{5}=-\frac{1}{5}i$$

Tenglamaning ikkala tarafidan \frac{3}{5} ni ayirish.

$$x=-\frac{3}{5}+\frac{1}{5}i x=-\frac{3}{5}-\frac{1}{5}i$$