5x^2+2x+8=0 eching

$$5 x ^ { 2 } + 2 x + 8 = 0$$

x uchun yechish (complex solution)

$$x=\frac{-1+\sqrt{39}i}{5}\approx -0,2+1,2489996i$$

$$x=\frac{-\sqrt{39}i-1}{5}\approx -0,2-1,2489996i$$

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.

$$5x^{2}+2x+8=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 2 ni b va 8 ni c bilan almashtiring.

$$x=\frac{-2±\sqrt{2^{2}-4\times 5\times 8}}{2\times 5}$$

2 kvadratini chiqarish.

$$x=\frac{-2±\sqrt{4-4\times 5\times 8}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4-20\times 8}}{2\times 5}$$

-20 ni 8 marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4-160}}{2\times 5}$$

4 ni -160 ga qo'shish.

$$x=\frac{-2±\sqrt{-156}}{2\times 5}$$

-156 ning kvadrat ildizini chiqarish.

$$x=\frac{-2±2\sqrt{39}i}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-2±2\sqrt{39}i}{10}$$

x=\frac{-2±2\sqrt{39}i}{10} tenglamasini yeching, bunda ± musbat. -2 ni 2i\sqrt{39} ga qo'shish.

$$x=\frac{-2+2\sqrt{39}i}{10}$$

-2+2i\sqrt{39} ni 10 ga bo'lish.

$$x=\frac{-1+\sqrt{39}i}{5}$$

x=\frac{-2±2\sqrt{39}i}{10} tenglamasini yeching, bunda ± manfiy. -2 dan 2i\sqrt{39} ni ayirish.

$$x=\frac{-2\sqrt{39}i-2}{10}$$

-2-2i\sqrt{39} ni 10 ga bo'lish.

$$x=\frac{-\sqrt{39}i-1}{5}$$

Tenglama yechildi.

$$x=\frac{-1+\sqrt{39}i}{5} x=\frac{-\sqrt{39}i-1}{5}$$

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.

$$5x^{2}+2x+8=0$$

Tenglamaning ikkala tarafidan 8 ni ayirish.

$$5x^{2}+2x+8-8=-8$$

O‘zidan 8 ayirilsa 0 qoladi.

$$5x^{2}+2x=-8$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}+2x}{5}=-\frac{8}{5}$$

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

$$x^{2}+\frac{2}{5}x=-\frac{8}{5}$$

\frac{2}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{5} olish uchun. Keyin, \frac{1}{5} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{2}{5}x+\left(\frac{1}{5}\right)^{2}=-\frac{8}{5}+\left(\frac{1}{5}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{5} kvadratini chiqarish.

$$x^{2}+\frac{2}{5}x+\frac{1}{25}=-\frac{8}{5}+\frac{1}{25}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{8}{5} ni \frac{1}{25} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{2}{5}x+\frac{1}{25}=-\frac{39}{25}$$

x^{2}+\frac{2}{5}x+\frac{1}{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.

$$\left(x+\frac{1}{5}\right)^{2}=-\frac{39}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{5}\right)^{2}}=\sqrt{-\frac{39}{25}}$$

Qisqartirish.

$$x+\frac{1}{5}=\frac{\sqrt{39}i}{5} x+\frac{1}{5}=-\frac{\sqrt{39}i}{5}$$

Tenglamaning ikkala tarafidan \frac{1}{5} ni ayirish.

$$x=\frac{-1+\sqrt{39}i}{5} x=\frac{-\sqrt{39}i-1}{5}$$