5x^2-2x+1=0 eching

$$5 x ^ { 2 } - 2 x + 1 = 0$$

x uchun yechish (complex solution)

$$x=\frac{1}{5}+\frac{2}{5}i=0,2+0,4i$$

$$x=\frac{1}{5}-\frac{2}{5}i=0,2-0,4i$$

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.

$$5x^{2}-2x+1=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, -2 ni b va 1 ni c bilan almashtiring.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 5}}{2\times 5}$$

-2 kvadratini chiqarish.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 5}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4-20}}{2\times 5}$$

4 ni -20 ga qo'shish.

$$x=\frac{-\left(-2\right)±\sqrt{-16}}{2\times 5}$$

-16 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-2\right)±4i}{2\times 5}$$

-2 ning teskarisi 2 ga teng.

$$x=\frac{2±4i}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{2±4i}{10}$$

x=\frac{2±4i}{10} tenglamasini yeching, bunda ± musbat. 2 ni 4i ga qo'shish.

$$x=\frac{2+4i}{10}$$

2+4i ni 10 ga bo'lish.

$$x=\frac{1}{5}+\frac{2}{5}i$$

x=\frac{2±4i}{10} tenglamasini yeching, bunda ± manfiy. 2 dan 4i ni ayirish.

$$x=\frac{2-4i}{10}$$

2-4i ni 10 ga bo'lish.

$$x=\frac{1}{5}-\frac{2}{5}i$$

Tenglama yechildi.

$$x=\frac{1}{5}+\frac{2}{5}i x=\frac{1}{5}-\frac{2}{5}i$$

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.

$$5x^{2}-2x+1=0$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$5x^{2}-2x+1-1=-1$$

O‘zidan 1 ayirilsa 0 qoladi.

$$5x^{2}-2x=-1$$

Ikki tarafini 5 ga bo‘ling.

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

5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.

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

-\frac{2}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{5} olish uchun. Keyin, -\frac{1}{5} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{5} kvadratini chiqarish.

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

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{1}{5} ni \frac{1}{25} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

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

x^{2}-\frac{2}{5}x+\frac{1}{25} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{5}\right)^{2}=-\frac{4}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{-\frac{4}{25}}$$

Qisqartirish.

$$x-\frac{1}{5}=\frac{2}{5}i x-\frac{1}{5}=-\frac{2}{5}i$$

\frac{1}{5} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{1}{5}+\frac{2}{5}i x=\frac{1}{5}-\frac{2}{5}i$$