3n^2+3n+1=1141 eching

$$3 n ^ { 2 } + 3 n + 1 = 1141$$

n uchun yechish

$$n=-20$$

$$n=19$$

Guruhlash orqali faktorlash qadamlari

Ikkala tarafdan 1141 ni ayirish.

$$3n^{2}+3n+1-1141=0$$

-1140 olish uchun 1 dan 1141 ni ayirish.

$$3n^{2}+3n-1140=0$$

Ikki tarafini 3 ga bo‘ling.

$$n^{2}+n-380=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon n^{2}+an+bn-380 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=1 ab=1\left(-380\right)=-380$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -380-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,380 -2,190 -4,95 -5,76 -10,38 -19,20$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+380=379 -2+190=188 -4+95=91 -5+76=71 -10+38=28 -19+20=1$$

Yechim – 1 yigʻindisini beruvchi juftlik.

$$a=-19 b=20$$

n^{2}+n-380 ni \left(n^{2}-19n\right)+\left(20n-380\right) sifatida qaytadan yozish.

$$\left(n^{2}-19n\right)+\left(20n-380\right)$$

Birinchi guruhda n ni va ikkinchi guruhda 20 ni faktordan chiqaring.

$$n\left(n-19\right)+20\left(n-19\right)$$

Distributiv funktsiyasidan foydalangan holda n-19 umumiy terminini chiqaring.

$$\left(n-19\right)\left(n+20\right)$$

Tenglamani yechish uchun n-19=0 va n+20=0 ni yeching.

$$n=19 n=-20$$

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.

$$3n^{2}+3n+1=1141$$

Tenglamaning ikkala tarafidan 1141 ni ayirish.

$$3n^{2}+3n+1-1141=1141-1141$$

O‘zidan 1141 ayirilsa 0 qoladi.

$$3n^{2}+3n+1-1141=0$$

1 dan 1141 ni ayirish.

$$3n^{2}+3n-1140=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, 3 ni b va -1140 ni c bilan almashtiring.

$$n=\frac{-3±\sqrt{3^{2}-4\times 3\left(-1140\right)}}{2\times 3}$$

3 kvadratini chiqarish.

$$n=\frac{-3±\sqrt{9-4\times 3\left(-1140\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$n=\frac{-3±\sqrt{9-12\left(-1140\right)}}{2\times 3}$$

-12 ni -1140 marotabaga ko'paytirish.

$$n=\frac{-3±\sqrt{9+13680}}{2\times 3}$$

9 ni 13680 ga qo'shish.

$$n=\frac{-3±\sqrt{13689}}{2\times 3}$$

13689 ning kvadrat ildizini chiqarish.

$$n=\frac{-3±117}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$n=\frac{-3±117}{6}$$

n=\frac{-3±117}{6} tenglamasini yeching, bunda ± musbat. -3 ni 117 ga qo'shish.

$$n=\frac{114}{6}$$

114 ni 6 ga bo'lish.

$$n=19$$

n=\frac{-3±117}{6} tenglamasini yeching, bunda ± manfiy. -3 dan 117 ni ayirish.

$$n=-\frac{120}{6}$$

-120 ni 6 ga bo'lish.

$$n=-20$$

Tenglama yechildi.

$$n=19 n=-20$$

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.

$$3n^{2}+3n+1=1141$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$3n^{2}+3n+1-1=1141-1$$

O‘zidan 1 ayirilsa 0 qoladi.

$$3n^{2}+3n=1141-1$$

1141 dan 1 ni ayirish.

$$3n^{2}+3n=1140$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3n^{2}+3n}{3}=\frac{1140}{3}$$

3 ga bo'lish 3 ga ko'paytirishni bekor qiladi.

$$n^{2}+\frac{3}{3}n=\frac{1140}{3}$$

3 ni 3 ga bo'lish.

$$n^{2}+n=\frac{1140}{3}$$

1140 ni 3 ga bo'lish.

$$n^{2}+n=380$$

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.

$$n^{2}+n+\left(\frac{1}{2}\right)^{2}=380+\left(\frac{1}{2}\right)^{2}$$

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

$$n^{2}+n+\frac{1}{4}=380+\frac{1}{4}$$

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

$$n^{2}+n+\frac{1}{4}=\frac{1521}{4}$$

n^{2}+n+\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.

$$\left(n+\frac{1}{2}\right)^{2}=\frac{1521}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(n+\frac{1}{2}\right)^{2}}=\sqrt{\frac{1521}{4}}$$

Qisqartirish.

$$n+\frac{1}{2}=\frac{39}{2} n+\frac{1}{2}=-\frac{39}{2}$$

Tenglamaning ikkala tarafidan \frac{1}{2} ni ayirish.

$$n=19 n=-20$$