2x^2+1-x=0 eching

$$2 { x }^{ 2 } +1-x=0$$

x uchun yechish (complex solution)

$$x=\frac{1+\sqrt{7}i}{4}\approx 0,25+0,661437828i$$

$$x=\frac{-\sqrt{7}i+1}{4}\approx 0,25-0,661437828i$$

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.

$$2x^{2}-x+1=0$$

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 1 ni c bilan almashtiring.

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

-4 ni 2 marotabaga ko'paytirish.

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

1 ni -8 ga qo'shish.

$$x=\frac{-\left(-1\right)±\sqrt{-7}}{2\times 2}$$

-7 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-1\right)±\sqrt{7}i}{2\times 2}$$

-1 ning teskarisi 1 ga teng.

$$x=\frac{1±\sqrt{7}i}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{1±\sqrt{7}i}{4}$$

x=\frac{1±\sqrt{7}i}{4} tenglamasini yeching, bunda ± musbat. 1 ni i\sqrt{7} ga qo'shish.

$$x=\frac{1+\sqrt{7}i}{4}$$

x=\frac{1±\sqrt{7}i}{4} tenglamasini yeching, bunda ± manfiy. 1 dan i\sqrt{7} ni ayirish.

$$x=\frac{-\sqrt{7}i+1}{4}$$

Tenglama yechildi.

$$x=\frac{1+\sqrt{7}i}{4} x=\frac{-\sqrt{7}i+1}{4}$$

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.

$$2x^{2}-x+1=0$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$2x^{2}-x+1-1=-1$$

O‘zidan 1 ayirilsa 0 qoladi.

$$2x^{2}-x=-1$$

Ikki tarafini 2 ga bo‘ling.

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

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

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

-\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.

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

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

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

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{1}{2} ni \frac{1}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{7}{16}$$

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.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{7}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{7}{16}}$$

Qisqartirish.

$$x-\frac{1}{4}=\frac{\sqrt{7}i}{4} x-\frac{1}{4}=-\frac{\sqrt{7}i}{4}$$

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

$$x=\frac{1+\sqrt{7}i}{4} x=\frac{-\sqrt{7}i+1}{4}$$