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

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

x uchun yechish (complex solution)

$$x=-\frac{3}{5}+\frac{4}{5}i=-0,6+0,8i$$

$$x=-\frac{3}{5}-\frac{4}{5}i=-0,6-0,8i$$

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.

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

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 5 ni c bilan almashtiring.

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

6 kvadratini chiqarish.

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

-4 ni 5 marotabaga ko'paytirish.

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

-20 ni 5 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-100}}{2\times 5}$$

36 ni -100 ga qo'shish.

$$x=\frac{-6±\sqrt{-64}}{2\times 5}$$

-64 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±8i}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-6±8i}{10}$$

x=\frac{-6±8i}{10} tenglamasini yeching, bunda ± musbat. -6 ni 8i ga qo'shish.

$$x=\frac{-6+8i}{10}$$

-6+8i ni 10 ga bo'lish.

$$x=-\frac{3}{5}+\frac{4}{5}i$$

x=\frac{-6±8i}{10} tenglamasini yeching, bunda ± manfiy. -6 dan 8i ni ayirish.

$$x=\frac{-6-8i}{10}$$

-6-8i ni 10 ga bo'lish.

$$x=-\frac{3}{5}-\frac{4}{5}i$$

Tenglama yechildi.

$$x=-\frac{3}{5}+\frac{4}{5}i x=-\frac{3}{5}-\frac{4}{5}i$$

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.

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

Tenglamaning ikkala tarafidan 5 ni ayirish.

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

O‘zidan 5 ayirilsa 0 qoladi.

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

Ikki tarafini 5 ga bo‘ling.

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

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

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

-5 ni 5 ga bo'lish.

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

\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.

$$x^{2}+\frac{6}{5}x+\left(\frac{3}{5}\right)^{2}=-1+\left(\frac{3}{5}\right)^{2}$$

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

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

-1 ni \frac{9}{25} ga qo'shish.

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=-\frac{16}{25}$$

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.

$$\left(x+\frac{3}{5}\right)^{2}=-\frac{16}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{5}\right)^{2}}=\sqrt{-\frac{16}{25}}$$

Qisqartirish.

$$x+\frac{3}{5}=\frac{4}{5}i x+\frac{3}{5}=-\frac{4}{5}i$$

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

$$x=-\frac{3}{5}+\frac{4}{5}i x=-\frac{3}{5}-\frac{4}{5}i$$