2x^2+6=x eching

$$2 { x }^{ 2 } +6=x$$

x uchun yechish (complex solution)

$$x=\frac{1+\sqrt{47}i}{4}\approx 0,25+1,71391365i$$

$$x=\frac{-\sqrt{47}i+1}{4}\approx 0,25-1,71391365i$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan x ni ayirish.

$$2x^{2}+6-x=0$$

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.

$$2x^{2}-x+6=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, -1 ni b va 6 ni c bilan almashtiring.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 2\times 6}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-1\right)±\sqrt{1-8\times 6}}{2\times 2}$$

-8 ni 6 marotabaga ko'paytirish.

$$x=\frac{-\left(-1\right)±\sqrt{1-48}}{2\times 2}$$

1 ni -48 ga qo'shish.

$$x=\frac{-\left(-1\right)±\sqrt{-47}}{2\times 2}$$

-47 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-1\right)±\sqrt{47}i}{2\times 2}$$

-1 ning teskarisi 1 ga teng.

$$x=\frac{1±\sqrt{47}i}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{1±\sqrt{47}i}{4}$$

x=\frac{1±\sqrt{47}i}{4} tenglamasini yeching, bunda ± musbat. 1 ni i\sqrt{47} ga qo'shish.

$$x=\frac{1+\sqrt{47}i}{4}$$

x=\frac{1±\sqrt{47}i}{4} tenglamasini yeching, bunda ± manfiy. 1 dan i\sqrt{47} ni ayirish.

$$x=\frac{-\sqrt{47}i+1}{4}$$

Tenglama yechildi.

$$x=\frac{1+\sqrt{47}i}{4} x=\frac{-\sqrt{47}i+1}{4}$$

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan x ni ayirish.

$$2x^{2}+6-x=0$$

Ikkala tarafdan 6 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

$$2x^{2}-x=-6$$

Ikki tarafini 2 ga bo‘ling.

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

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}-\frac{1}{2}x=-\frac{6}{2}$$

-6 ni 2 ga bo'lish.

$$x^{2}-\frac{1}{2}x=-3$$

-\frac{1}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{4} olish uchun. Keyin, -\frac{1}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=-3+\left(-\frac{1}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{4} kvadratini chiqarish.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-3+\frac{1}{16}$$

-3 ni \frac{1}{16} ga qo'shish.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{47}{16}$$

x^{2}-\frac{1}{2}x+\frac{1}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{47}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{47}{16}}$$

Qisqartirish.

$$x-\frac{1}{4}=\frac{\sqrt{47}i}{4} x-\frac{1}{4}=-\frac{\sqrt{47}i}{4}$$

\frac{1}{4} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{1+\sqrt{47}i}{4} x=\frac{-\sqrt{47}i+1}{4}$$