2x^2-2x=1 eching

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

x uchun yechish

$$x = \frac{\sqrt{3} + 1}{2} \approx 1,366025404$$

$$x=\frac{1-\sqrt{3}}{2}\approx -0,366025404$$

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.

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

Tenglamaning ikkala tarafidan 1 ni ayirish.

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

O‘zidan 1 ayirilsa 0 qoladi.

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

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

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

-2 kvadratini chiqarish.

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

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4-8\left(-1\right)}}{2\times 2}$$

-8 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4+8}}{2\times 2}$$

4 ni 8 ga qo'shish.

$$x=\frac{-\left(-2\right)±\sqrt{12}}{2\times 2}$$

12 ning kvadrat ildizini chiqarish.

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

-2 ning teskarisi 2 ga teng.

$$x=\frac{2±2\sqrt{3}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{2±2\sqrt{3}}{4}$$

x=\frac{2±2\sqrt{3}}{4} tenglamasini yeching, bunda ± musbat. 2 ni 2\sqrt{3} ga qo'shish.

$$x=\frac{2\sqrt{3}+2}{4}$$

2+2\sqrt{3} ni 4 ga bo'lish.

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

x=\frac{2±2\sqrt{3}}{4} tenglamasini yeching, bunda ± manfiy. 2 dan 2\sqrt{3} ni ayirish.

$$x=\frac{2-2\sqrt{3}}{4}$$

2-2\sqrt{3} ni 4 ga bo'lish.

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

Tenglama yechildi.

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

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.

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

Ikki tarafini 2 ga bo‘ling.

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

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

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

-2 ni 2 ga bo'lish.

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

-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.

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

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

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

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

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

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.

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

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

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