2x^2-6x+1=0 eching

$$2 x ^ { 2 } - 6 x + 1 = 0$$

x uchun yechish

$$x = \frac{\sqrt{7} + 3}{2} \approx 2,822875656$$

$$x=\frac{3-\sqrt{7}}{2}\approx 0,177124344$$

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.

$$2x^{2}-6x+1=0$$

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

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

-6 kvadratini chiqarish.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 2}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-6\right)±\sqrt{36-8}}{2\times 2}$$

36 ni -8 ga qo'shish.

$$x=\frac{-\left(-6\right)±\sqrt{28}}{2\times 2}$$

28 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-6\right)±2\sqrt{7}}{2\times 2}$$

-6 ning teskarisi 6 ga teng.

$$x=\frac{6±2\sqrt{7}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{6±2\sqrt{7}}{4}$$

x=\frac{6±2\sqrt{7}}{4} tenglamasini yeching, bunda ± musbat. 6 ni 2\sqrt{7} ga qo'shish.

$$x=\frac{2\sqrt{7}+6}{4}$$

6+2\sqrt{7} ni 4 ga bo'lish.

$$x=\frac{\sqrt{7}+3}{2}$$

x=\frac{6±2\sqrt{7}}{4} tenglamasini yeching, bunda ± manfiy. 6 dan 2\sqrt{7} ni ayirish.

$$x=\frac{6-2\sqrt{7}}{4}$$

6-2\sqrt{7} ni 4 ga bo'lish.

$$x=\frac{3-\sqrt{7}}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{7}+3}{2} x=\frac{3-\sqrt{7}}{2}$$

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.

$$2x^{2}-6x+1=0$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$2x^{2}-6x+1-1=-1$$

O‘zidan 1 ayirilsa 0 qoladi.

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

Ikki tarafini 2 ga bo‘ling.

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

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

$$x^{2}+\left(-\frac{6}{2}\right)x=-\frac{1}{2}$$

-6 ni 2 ga bo'lish.

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

-3 ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{3}{2} olish uchun. Keyin, -\frac{3}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{3}{2} kvadratini chiqarish.

$$x^{2}-3x+\frac{9}{4}=-\frac{1}{2}+\frac{9}{4}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{1}{2} ni \frac{9}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-3x+\frac{9}{4}=\frac{7}{4}$$

x^{2}-3x+\frac{9}{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.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{7}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{7}{4}}$$

Qisqartirish.

$$x-\frac{3}{2}=\frac{\sqrt{7}}{2} x-\frac{3}{2}=-\frac{\sqrt{7}}{2}$$

\frac{3}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{7}+3}{2} x=\frac{3-\sqrt{7}}{2}$$