2(x+1)^2-1=71 eching

$$2 ( x + 1 ) ^ { 2 } - 1 = 71$$

x uchun yechish

$$x=5$$

$$x=-7$$

Guruhlash orqali faktorlash qadamlari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+1\right)^{2} kengaytirilishi uchun ishlating.

$$2\left(x^{2}+2x+1\right)-1=71$$

2 ga x^{2}+2x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2x^{2}+4x+2-1=71$$

1 olish uchun 2 dan 1 ni ayirish.

$$2x^{2}+4x+1=71$$

Ikkala tarafdan 71 ni ayirish.

$$2x^{2}+4x+1-71=0$$

-70 olish uchun 1 dan 71 ni ayirish.

$$2x^{2}+4x-70=0$$

Ikki tarafini 2 ga bo‘ling.

$$x^{2}+2x-35=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon x^{2}+ax+bx-35 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=2 ab=1\left(-35\right)=-35$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -35-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,35 -5,7$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+35=34 -5+7=2$$

Yechim – 2 yigʻindisini beruvchi juftlik.

$$a=-5 b=7$$

x^{2}+2x-35 ni \left(x^{2}-5x\right)+\left(7x-35\right) sifatida qaytadan yozish.

$$\left(x^{2}-5x\right)+\left(7x-35\right)$$

Birinchi guruhda x ni va ikkinchi guruhda 7 ni faktordan chiqaring.

$$x\left(x-5\right)+7\left(x-5\right)$$

Distributiv funktsiyasidan foydalangan holda x-5 umumiy terminini chiqaring.

$$\left(x-5\right)\left(x+7\right)$$

Tenglamani yechish uchun x-5=0 va x+7=0 ni yeching.

$$x=5 x=-7$$

Kvadrat tenglamalaridan foydalanish qoidalari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+1\right)^{2} kengaytirilishi uchun ishlating.

$$2\left(x^{2}+2x+1\right)-1=71$$

2 ga x^{2}+2x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2x^{2}+4x+2-1=71$$

1 olish uchun 2 dan 1 ni ayirish.

$$2x^{2}+4x+1=71$$

Ikkala tarafdan 71 ni ayirish.

$$2x^{2}+4x+1-71=0$$

-70 olish uchun 1 dan 71 ni ayirish.

$$2x^{2}+4x-70=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, 4 ni b va -70 ni c bilan almashtiring.

$$x=\frac{-4±\sqrt{4^{2}-4\times 2\left(-70\right)}}{2\times 2}$$

4 kvadratini chiqarish.

$$x=\frac{-4±\sqrt{16-4\times 2\left(-70\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16-8\left(-70\right)}}{2\times 2}$$

-8 ni -70 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16+560}}{2\times 2}$$

16 ni 560 ga qo'shish.

$$x=\frac{-4±\sqrt{576}}{2\times 2}$$

576 ning kvadrat ildizini chiqarish.

$$x=\frac{-4±24}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-4±24}{4}$$

x=\frac{-4±24}{4} tenglamasini yeching, bunda ± musbat. -4 ni 24 ga qo'shish.

$$x=\frac{20}{4}$$

20 ni 4 ga bo'lish.

$$x=5$$

x=\frac{-4±24}{4} tenglamasini yeching, bunda ± manfiy. -4 dan 24 ni ayirish.

$$x=-\frac{28}{4}$$

-28 ni 4 ga bo'lish.

$$x=-7$$

Tenglama yechildi.

$$x=5 x=-7$$

Kvadratni yakunlash bo'yicha qadamlar

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+1\right)^{2} kengaytirilishi uchun ishlating.

$$2\left(x^{2}+2x+1\right)-1=71$$

2 ga x^{2}+2x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2x^{2}+4x+2-1=71$$

1 olish uchun 2 dan 1 ni ayirish.

$$2x^{2}+4x+1=71$$

Ikkala tarafdan 1 ni ayirish.

$$2x^{2}+4x=71-1$$

70 olish uchun 71 dan 1 ni ayirish.

$$2x^{2}+4x=70$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+4x}{2}=\frac{70}{2}$$

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

$$x^{2}+\frac{4}{2}x=\frac{70}{2}$$

4 ni 2 ga bo'lish.

$$x^{2}+2x=\frac{70}{2}$$

70 ni 2 ga bo'lish.

$$x^{2}+2x=35$$

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

$$x^{2}+2x+1^{2}=35+1^{2}$$

1 kvadratini chiqarish.

$$x^{2}+2x+1=35+1$$

35 ni 1 ga qo'shish.

$$x^{2}+2x+1=36$$

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

$$\left(x+1\right)^{2}=36$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{36}$$

Qisqartirish.

$$x+1=6 x+1=-6$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x=5 x=-7$$