4x^2+8=-5x eching

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

x uchun yechish (complex solution)

$$x=\frac{-5+\sqrt{103}i}{8}\approx -0,625+1,268611446i$$

$$x=\frac{-\sqrt{103}i-5}{8}\approx -0,625-1,268611446i$$

Kvadrat tenglamalaridan foydalanish qoidalari

5x ni ikki tarafga qo’shing.

$$4x^{2}+8+5x=0$$

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.

$$4x^{2}+5x+8=0$$

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

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

5 kvadratini chiqarish.

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

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25-16\times 8}}{2\times 4}$$

-16 ni 8 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25-128}}{2\times 4}$$

25 ni -128 ga qo'shish.

$$x=\frac{-5±\sqrt{-103}}{2\times 4}$$

-103 ning kvadrat ildizini chiqarish.

$$x=\frac{-5±\sqrt{103}i}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{103}i}{8}$$

x=\frac{-5±\sqrt{103}i}{8} tenglamasini yeching, bunda ± musbat. -5 ni i\sqrt{103} ga qo'shish.

$$x=\frac{-5+\sqrt{103}i}{8}$$

x=\frac{-5±\sqrt{103}i}{8} tenglamasini yeching, bunda ± manfiy. -5 dan i\sqrt{103} ni ayirish.

$$x=\frac{-\sqrt{103}i-5}{8}$$

Tenglama yechildi.

$$x=\frac{-5+\sqrt{103}i}{8} x=\frac{-\sqrt{103}i-5}{8}$$

Kvadratni yakunlash bo'yicha qadamlar

5x ni ikki tarafga qo’shing.

$$4x^{2}+8+5x=0$$

Ikkala tarafdan 8 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

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

Ikki tarafini 4 ga bo‘ling.

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

4 ga bo'lish 4 ga ko'paytirishni bekor qiladi.

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

-8 ni 4 ga bo'lish.

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

\frac{5}{4} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{5}{8} olish uchun. Keyin, \frac{5}{8} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{5}{8} kvadratini chiqarish.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=-2+\frac{25}{64}$$

-2 ni \frac{25}{64} ga qo'shish.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=-\frac{103}{64}$$

x^{2}+\frac{5}{4}x+\frac{25}{64} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{5}{8}\right)^{2}=-\frac{103}{64}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{5}{8}\right)^{2}}=\sqrt{-\frac{103}{64}}$$

Qisqartirish.

$$x+\frac{5}{8}=\frac{\sqrt{103}i}{8} x+\frac{5}{8}=-\frac{\sqrt{103}i}{8}$$

Tenglamaning ikkala tarafidan \frac{5}{8} ni ayirish.

$$x=\frac{-5+\sqrt{103}i}{8} x=\frac{-\sqrt{103}i-5}{8}$$