x^2-x+1=0 eching

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

x uchun yechish (complex solution)

$$x=\frac{1+\sqrt{3}i}{2}\approx 0,5+0,866025404i$$

$$x=\frac{-\sqrt{3}i+1}{2}\approx 0,5-0,866025404i$$

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.

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

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

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

1 ni -4 ga qo'shish.

$$x=\frac{-\left(-1\right)±\sqrt{-3}}{2}$$

-3 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-1\right)±\sqrt{3}i}{2}$$

-1 ning teskarisi 1 ga teng.

$$x=\frac{1±\sqrt{3}i}{2}$$

x=\frac{1±\sqrt{3}i}{2} tenglamasini yeching, bunda ± musbat. 1 ni i\sqrt{3} ga qo'shish.

$$x=\frac{1+\sqrt{3}i}{2}$$

x=\frac{1±\sqrt{3}i}{2} tenglamasini yeching, bunda ± manfiy. 1 dan i\sqrt{3} ni ayirish.

$$x=\frac{-\sqrt{3}i+1}{2}$$

Tenglama yechildi.

$$x=\frac{1+\sqrt{3}i}{2} x=\frac{-\sqrt{3}i+1}{2}$$

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.

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

Tenglamaning ikkala tarafidan 1 ni ayirish.

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

O‘zidan 1 ayirilsa 0 qoladi.

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

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{2} kvadratini chiqarish.

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

-1 ni \frac{1}{4} ga qo'shish.

$$x^{2}-x+\frac{1}{4}=-\frac{3}{4}$$

x^{2}-x+\frac{1}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x-\frac{1}{2}=\frac{\sqrt{3}i}{2} x-\frac{1}{2}=-\frac{\sqrt{3}i}{2}$$

\frac{1}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{1+\sqrt{3}i}{2} x=\frac{-\sqrt{3}i+1}{2}$$