5x^2+4x+3=0 eching

$$5 x ^ { 2 } + 4 x + 3 = 0$$

x uchun yechish (complex solution)

$$x=\frac{-2+\sqrt{11}i}{5}\approx -0,4+0,663324958i$$

$$x=\frac{-\sqrt{11}i-2}{5}\approx -0,4-0,663324958i$$

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.

$$5x^{2}+4x+3=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 4 ni b va 3 ni c bilan almashtiring.

$$x=\frac{-4±\sqrt{4^{2}-4\times 5\times 3}}{2\times 5}$$

4 kvadratini chiqarish.

$$x=\frac{-4±\sqrt{16-4\times 5\times 3}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16-20\times 3}}{2\times 5}$$

-20 ni 3 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16-60}}{2\times 5}$$

16 ni -60 ga qo'shish.

$$x=\frac{-4±\sqrt{-44}}{2\times 5}$$

-44 ning kvadrat ildizini chiqarish.

$$x=\frac{-4±2\sqrt{11}i}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-4±2\sqrt{11}i}{10}$$

x=\frac{-4±2\sqrt{11}i}{10} tenglamasini yeching, bunda ± musbat. -4 ni 2i\sqrt{11} ga qo'shish.

$$x=\frac{-4+2\sqrt{11}i}{10}$$

-4+2i\sqrt{11} ni 10 ga bo'lish.

$$x=\frac{-2+\sqrt{11}i}{5}$$

x=\frac{-4±2\sqrt{11}i}{10} tenglamasini yeching, bunda ± manfiy. -4 dan 2i\sqrt{11} ni ayirish.

$$x=\frac{-2\sqrt{11}i-4}{10}$$

-4-2i\sqrt{11} ni 10 ga bo'lish.

$$x=\frac{-\sqrt{11}i-2}{5}$$

Tenglama yechildi.

$$x=\frac{-2+\sqrt{11}i}{5} x=\frac{-\sqrt{11}i-2}{5}$$

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.

$$5x^{2}+4x+3=0$$

Tenglamaning ikkala tarafidan 3 ni ayirish.

$$5x^{2}+4x+3-3=-3$$

O‘zidan 3 ayirilsa 0 qoladi.

$$5x^{2}+4x=-3$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}+4x}{5}=-\frac{3}{5}$$

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

$$x^{2}+\frac{4}{5}x=-\frac{3}{5}$$

\frac{4}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{2}{5} olish uchun. Keyin, \frac{2}{5} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{4}{5}x+\left(\frac{2}{5}\right)^{2}=-\frac{3}{5}+\left(\frac{2}{5}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{2}{5} kvadratini chiqarish.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=-\frac{3}{5}+\frac{4}{25}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{3}{5} ni \frac{4}{25} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=-\frac{11}{25}$$

x^{2}+\frac{4}{5}x+\frac{4}{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.

$$\left(x+\frac{2}{5}\right)^{2}=-\frac{11}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{2}{5}\right)^{2}}=\sqrt{-\frac{11}{25}}$$

Qisqartirish.

$$x+\frac{2}{5}=\frac{\sqrt{11}i}{5} x+\frac{2}{5}=-\frac{\sqrt{11}i}{5}$$

Tenglamaning ikkala tarafidan \frac{2}{5} ni ayirish.

$$x=\frac{-2+\sqrt{11}i}{5} x=\frac{-\sqrt{11}i-2}{5}$$