2x^2-2x-4=0 eching

$$2 x ^ { 2 } - 2 x - 4 = 0$$

x uchun yechish

$$x=-1$$

$$x=2$$

Guruhlash orqali faktorlash qadamlari

Ikki tarafini 2 ga bo‘ling.

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

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon x^{2}+ax+bx-2 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-1 ab=1\left(-2\right)=-2$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. Faqat bundan juftlik tizim yechimidir.

$$a=-2 b=1$$

x^{2}-x-2 ni \left(x^{2}-2x\right)+\left(x-2\right) sifatida qaytadan yozish.

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

x^{2}-2x ichida x ni ajrating.

$$x\left(x-2\right)+x-2$$

Distributiv funktsiyasidan foydalangan holda x-2 umumiy terminini chiqaring.

$$\left(x-2\right)\left(x+1\right)$$

Tenglamani yechish uchun x-2=0 va x+1=0 ni yeching.

$$x=2 x=-1$$

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.

$$2x^{2}-2x-4=0$$

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 -4 ni c bilan almashtiring.

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

-2 kvadratini chiqarish.

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

-4 ni 2 marotabaga ko'paytirish.

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

-8 ni -4 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4+32}}{2\times 2}$$

4 ni 32 ga qo'shish.

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

36 ning kvadrat ildizini chiqarish.

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

-2 ning teskarisi 2 ga teng.

$$x=\frac{2±6}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{2±6}{4}$$

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

$$x=\frac{8}{4}$$

8 ni 4 ga bo'lish.

$$x=2$$

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

$$x=-\frac{4}{4}$$

-4 ni 4 ga bo'lish.

$$x=-1$$

Tenglama yechildi.

$$x=2 x=-1$$

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.

$$2x^{2}-2x-4=0$$

4 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -4 ayirilsa 0 qoladi.

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

0 dan -4 ni ayirish.

$$2x^{2}-2x=4$$

Ikki tarafini 2 ga bo‘ling.

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

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

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

-2 ni 2 ga bo'lish.

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

4 ni 2 ga bo'lish.

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

-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.

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

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

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

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

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

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.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Qisqartirish.

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

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

$$x=2 x=-1$$