5x^2+6x+10=0 eching

$$5 x ^ { 2 } + 6 x + 10 = 0$$

x uchun yechish (complex solution)

$$x=\frac{-3+\sqrt{41}i}{5}\approx -0,6+1,280624847i$$

$$x=\frac{-\sqrt{41}i-3}{5}\approx -0,6-1,280624847i$$

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.

$$5x^{2}+6x+10=0$$

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 10 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}-4\times 5\times 10}}{2\times 5}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\times 5\times 10}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-20\times 10}}{2\times 5}$$

-20 ni 10 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-200}}{2\times 5}$$

36 ni -200 ga qo'shish.

$$x=\frac{-6±\sqrt{-164}}{2\times 5}$$

-164 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±2\sqrt{41}i}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-6±2\sqrt{41}i}{10}$$

x=\frac{-6±2\sqrt{41}i}{10} tenglamasini yeching, bunda ± musbat. -6 ni 2i\sqrt{41} ga qo'shish.

$$x=\frac{-6+2\sqrt{41}i}{10}$$

-6+2i\sqrt{41} ni 10 ga bo'lish.

$$x=\frac{-3+\sqrt{41}i}{5}$$

x=\frac{-6±2\sqrt{41}i}{10} tenglamasini yeching, bunda ± manfiy. -6 dan 2i\sqrt{41} ni ayirish.

$$x=\frac{-2\sqrt{41}i-6}{10}$$

-6-2i\sqrt{41} ni 10 ga bo'lish.

$$x=\frac{-\sqrt{41}i-3}{5}$$

Tenglama yechildi.

$$x=\frac{-3+\sqrt{41}i}{5} x=\frac{-\sqrt{41}i-3}{5}$$

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.

$$5x^{2}+6x+10=0$$

Tenglamaning ikkala tarafidan 10 ni ayirish.

$$5x^{2}+6x+10-10=-10$$

O‘zidan 10 ayirilsa 0 qoladi.

$$5x^{2}+6x=-10$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}+6x}{5}=-\frac{10}{5}$$

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

$$x^{2}+\frac{6}{5}x=-\frac{10}{5}$$

-10 ni 5 ga bo'lish.

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

\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.

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

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

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

-2 ni \frac{9}{25} ga qo'shish.

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=-\frac{41}{25}$$

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.

$$\left(x+\frac{3}{5}\right)^{2}=-\frac{41}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{5}\right)^{2}}=\sqrt{-\frac{41}{25}}$$

Qisqartirish.

$$x+\frac{3}{5}=\frac{\sqrt{41}i}{5} x+\frac{3}{5}=-\frac{\sqrt{41}i}{5}$$

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

$$x=\frac{-3+\sqrt{41}i}{5} x=\frac{-\sqrt{41}i-3}{5}$$