3n^2-11n-874=0 eching

$$3 { n }^{ 2 } -11n-874 = 0$$

n uchun yechish

$$n = -\frac{46}{3} = -15\frac{1}{3} \approx -15,333333333$$

$$n=19$$

Guruhlash orqali faktorlash qadamlari

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 3n^{2}+an+bn-874 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-11 ab=3\left(-874\right)=-2622$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -2622-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-2622 2,-1311 3,-874 6,-437 19,-138 23,-114 38,-69 46,-57$$

Har bir juftlik yigʻindisini hisoblang.

$$1-2622=-2621 2-1311=-1309 3-874=-871 6-437=-431 19-138=-119 23-114=-91 38-69=-31 46-57=-11$$

Yechim – -11 yigʻindisini beruvchi juftlik.

$$a=-57 b=46$$

3n^{2}-11n-874 ni \left(3n^{2}-57n\right)+\left(46n-874\right) sifatida qaytadan yozish.

$$\left(3n^{2}-57n\right)+\left(46n-874\right)$$

Birinchi guruhda 3n ni va ikkinchi guruhda 46 ni faktordan chiqaring.

$$3n\left(n-19\right)+46\left(n-19\right)$$

Distributiv funktsiyasidan foydalangan holda n-19 umumiy terminini chiqaring.

$$\left(n-19\right)\left(3n+46\right)$$

Tenglamani yechish uchun n-19=0 va 3n+46=0 ni yeching.

$$n=19 n=-\frac{46}{3}$$

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.

$$3n^{2}-11n-874=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, -11 ni b va -874 ni c bilan almashtiring.

$$n=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 3\left(-874\right)}}{2\times 3}$$

-11 kvadratini chiqarish.

$$n=\frac{-\left(-11\right)±\sqrt{121-4\times 3\left(-874\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$n=\frac{-\left(-11\right)±\sqrt{121-12\left(-874\right)}}{2\times 3}$$

-12 ni -874 marotabaga ko'paytirish.

$$n=\frac{-\left(-11\right)±\sqrt{121+10488}}{2\times 3}$$

121 ni 10488 ga qo'shish.

$$n=\frac{-\left(-11\right)±\sqrt{10609}}{2\times 3}$$

10609 ning kvadrat ildizini chiqarish.

$$n=\frac{-\left(-11\right)±103}{2\times 3}$$

-11 ning teskarisi 11 ga teng.

$$n=\frac{11±103}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$n=\frac{11±103}{6}$$

n=\frac{11±103}{6} tenglamasini yeching, bunda ± musbat. 11 ni 103 ga qo'shish.

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

114 ni 6 ga bo'lish.

$$n=19$$

n=\frac{11±103}{6} tenglamasini yeching, bunda ± manfiy. 11 dan 103 ni ayirish.

$$n=-\frac{92}{6}$$

\frac{-92}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$n=-\frac{46}{3}$$

Tenglama yechildi.

$$n=19 n=-\frac{46}{3}$$

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.

$$3n^{2}-11n-874=0$$

874 ni tenglamaning ikkala tarafiga qo'shish.

$$3n^{2}-11n-874-\left(-874\right)=-\left(-874\right)$$

O‘zidan -874 ayirilsa 0 qoladi.

$$3n^{2}-11n=-\left(-874\right)$$

0 dan -874 ni ayirish.

$$3n^{2}-11n=874$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3n^{2}-11n}{3}=\frac{874}{3}$$

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

$$n^{2}-\frac{11}{3}n=\frac{874}{3}$$

-\frac{11}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{11}{6} olish uchun. Keyin, -\frac{11}{6} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$n^{2}-\frac{11}{3}n+\left(-\frac{11}{6}\right)^{2}=\frac{874}{3}+\left(-\frac{11}{6}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{11}{6} kvadratini chiqarish.

$$n^{2}-\frac{11}{3}n+\frac{121}{36}=\frac{874}{3}+\frac{121}{36}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{874}{3} ni \frac{121}{36} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$n^{2}-\frac{11}{3}n+\frac{121}{36}=\frac{10609}{36}$$

n^{2}-\frac{11}{3}n+\frac{121}{36} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(n-\frac{11}{6}\right)^{2}=\frac{10609}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(n-\frac{11}{6}\right)^{2}}=\sqrt{\frac{10609}{36}}$$

Qisqartirish.

$$n-\frac{11}{6}=\frac{103}{6} n-\frac{11}{6}=-\frac{103}{6}$$

\frac{11}{6} ni tenglamaning ikkala tarafiga qo'shish.

$$n=19 n=-\frac{46}{3}$$