4(x+1)^2-169=0 eching

$$4 ( x + 1 ) ^ { 2 } - 169 = 0$$

x uchun yechish

$$x = \frac{11}{2} = 5\frac{1}{2} = 5,5$$

$$x = -\frac{15}{2} = -7\frac{1}{2} = -7,5$$

Guruhlash orqali faktorlash qadamlari

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

$$4\left(x^{2}+2x+1\right)-169=0$$

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

$$4x^{2}+8x+4-169=0$$

-165 olish uchun 4 dan 169 ni ayirish.

$$4x^{2}+8x-165=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 4x^{2}+ax+bx-165 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=8 ab=4\left(-165\right)=-660$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -660-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,660 -2,330 -3,220 -4,165 -5,132 -6,110 -10,66 -11,60 -12,55 -15,44 -20,33 -22,30$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+660=659 -2+330=328 -3+220=217 -4+165=161 -5+132=127 -6+110=104 -10+66=56 -11+60=49 -12+55=43 -15+44=29 -20+33=13 -22+30=8$$

Yechim – 8 yigʻindisini beruvchi juftlik.

$$a=-22 b=30$$

4x^{2}+8x-165 ni \left(4x^{2}-22x\right)+\left(30x-165\right) sifatida qaytadan yozish.

$$\left(4x^{2}-22x\right)+\left(30x-165\right)$$

Birinchi guruhda 2x ni va ikkinchi guruhda 15 ni faktordan chiqaring.

$$2x\left(2x-11\right)+15\left(2x-11\right)$$

Distributiv funktsiyasidan foydalangan holda 2x-11 umumiy terminini chiqaring.

$$\left(2x-11\right)\left(2x+15\right)$$

Tenglamani yechish uchun 2x-11=0 va 2x+15=0 ni yeching.

$$x=\frac{11}{2} x=-\frac{15}{2}$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

$$4\left(x^{2}+2x+1\right)-169=0$$

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

$$4x^{2}+8x+4-169=0$$

-165 olish uchun 4 dan 169 ni ayirish.

$$4x^{2}+8x-165=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 4 ni a, 8 ni b va -165 ni c bilan almashtiring.

$$x=\frac{-8±\sqrt{8^{2}-4\times 4\left(-165\right)}}{2\times 4}$$

8 kvadratini chiqarish.

$$x=\frac{-8±\sqrt{64-4\times 4\left(-165\right)}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64-16\left(-165\right)}}{2\times 4}$$

-16 ni -165 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64+2640}}{2\times 4}$$

64 ni 2640 ga qo'shish.

$$x=\frac{-8±\sqrt{2704}}{2\times 4}$$

2704 ning kvadrat ildizini chiqarish.

$$x=\frac{-8±52}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{-8±52}{8}$$

x=\frac{-8±52}{8} tenglamasini yeching, bunda ± musbat. -8 ni 52 ga qo'shish.

$$x=\frac{44}{8}$$

\frac{44}{8} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{11}{2}$$

x=\frac{-8±52}{8} tenglamasini yeching, bunda ± manfiy. -8 dan 52 ni ayirish.

$$x=-\frac{60}{8}$$

\frac{-60}{8} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{15}{2}$$

Tenglama yechildi.

$$x=\frac{11}{2} x=-\frac{15}{2}$$

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.

$$4\left(x^{2}+2x+1\right)-169=0$$

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

$$4x^{2}+8x+4-169=0$$

-165 olish uchun 4 dan 169 ni ayirish.

$$4x^{2}+8x-165=0$$

165 ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

$$4x^{2}+8x=165$$

Ikki tarafini 4 ga bo‘ling.

$$\frac{4x^{2}+8x}{4}=\frac{165}{4}$$

4 ga bo'lish 4 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{8}{4}x=\frac{165}{4}$$

8 ni 4 ga bo'lish.

$$x^{2}+2x=\frac{165}{4}$$

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.

$$x^{2}+2x+1^{2}=\frac{165}{4}+1^{2}$$

1 kvadratini chiqarish.

$$x^{2}+2x+1=\frac{165}{4}+1$$

\frac{165}{4} ni 1 ga qo'shish.

$$x^{2}+2x+1=\frac{169}{4}$$

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.

$$\left(x+1\right)^{2}=\frac{169}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{169}{4}}$$

Qisqartirish.

$$x+1=\frac{13}{2} x+1=-\frac{13}{2}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x=\frac{11}{2} x=-\frac{15}{2}$$