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

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

x uchun yechish (complex solution)

$$x=\frac{-5+\sqrt{39}i}{16}\approx -0,3125+0,390312375i$$

$$x=\frac{-\sqrt{39}i-5}{16}\approx -0,3125-0,390312375i$$

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.

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

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

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

5 kvadratini chiqarish.

$$x=\frac{-5±\sqrt{25-4\times 8\times 2}}{2\times 8}$$

-4 ni 8 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25-32\times 2}}{2\times 8}$$

-32 ni 2 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25-64}}{2\times 8}$$

25 ni -64 ga qo'shish.

$$x=\frac{-5±\sqrt{-39}}{2\times 8}$$

-39 ning kvadrat ildizini chiqarish.

$$x=\frac{-5±\sqrt{39}i}{2\times 8}$$

2 ni 8 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{39}i}{16}$$

x=\frac{-5±\sqrt{39}i}{16} tenglamasini yeching, bunda ± musbat. -5 ni i\sqrt{39} ga qo'shish.

$$x=\frac{-5+\sqrt{39}i}{16}$$

x=\frac{-5±\sqrt{39}i}{16} tenglamasini yeching, bunda ± manfiy. -5 dan i\sqrt{39} ni ayirish.

$$x=\frac{-\sqrt{39}i-5}{16}$$

Tenglama yechildi.

$$x=\frac{-5+\sqrt{39}i}{16} x=\frac{-\sqrt{39}i-5}{16}$$

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.

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

Tenglamaning ikkala tarafidan 2 ni ayirish.

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

O‘zidan 2 ayirilsa 0 qoladi.

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

Ikki tarafini 8 ga bo‘ling.

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

8 ga bo'lish 8 ga ko'paytirishni bekor qiladi.

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

\frac{-2}{8} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{5}{8}x=-\frac{1}{4}$$

\frac{5}{8} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{5}{16} olish uchun. Keyin, \frac{5}{16} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{5}{8}x+\left(\frac{5}{16}\right)^{2}=-\frac{1}{4}+\left(\frac{5}{16}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{5}{16} kvadratini chiqarish.

$$x^{2}+\frac{5}{8}x+\frac{25}{256}=-\frac{1}{4}+\frac{25}{256}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{1}{4} ni \frac{25}{256} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{5}{8}x+\frac{25}{256}=-\frac{39}{256}$$

x^{2}+\frac{5}{8}x+\frac{25}{256} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{5}{16}\right)^{2}=-\frac{39}{256}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{5}{16}\right)^{2}}=\sqrt{-\frac{39}{256}}$$

Qisqartirish.

$$x+\frac{5}{16}=\frac{\sqrt{39}i}{16} x+\frac{5}{16}=-\frac{\sqrt{39}i}{16}$$

Tenglamaning ikkala tarafidan \frac{5}{16} ni ayirish.

$$x=\frac{-5+\sqrt{39}i}{16} x=\frac{-\sqrt{39}i-5}{16}$$