9n^2-23n+20=3n^2 eching

$$9 n ^ { 2 } - 23 n + 20 = 3 n ^ { 2 }$$

n uchun yechish

$$n = \frac{4}{3} = 1\frac{1}{3} \approx 1,333333333$$

$$n = \frac{5}{2} = 2\frac{1}{2} = 2,5$$

Guruhlash orqali faktorlash qadamlari

Ikkala tarafdan 3n^{2} ni ayirish.

$$9n^{2}-23n+20-3n^{2}=0$$

6n^{2} ni olish uchun 9n^{2} va -3n^{2} ni birlashtirish.

$$6n^{2}-23n+20=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 6n^{2}+an+bn+20 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-23 ab=6\times 20=120$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 120-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-120 -2,-60 -3,-40 -4,-30 -5,-24 -6,-20 -8,-15 -10,-12$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-120=-121 -2-60=-62 -3-40=-43 -4-30=-34 -5-24=-29 -6-20=-26 -8-15=-23 -10-12=-22$$

Yechim – -23 yigʻindisini beruvchi juftlik.

$$a=-15 b=-8$$

6n^{2}-23n+20 ni \left(6n^{2}-15n\right)+\left(-8n+20\right) sifatida qaytadan yozish.

$$\left(6n^{2}-15n\right)+\left(-8n+20\right)$$

Birinchi guruhda 3n ni va ikkinchi guruhda -4 ni faktordan chiqaring.

$$3n\left(2n-5\right)-4\left(2n-5\right)$$

Distributiv funktsiyasidan foydalangan holda 2n-5 umumiy terminini chiqaring.

$$\left(2n-5\right)\left(3n-4\right)$$

Tenglamani yechish uchun 2n-5=0 va 3n-4=0 ni yeching.

$$n=\frac{5}{2} n=\frac{4}{3}$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan 3n^{2} ni ayirish.

$$9n^{2}-23n+20-3n^{2}=0$$

6n^{2} ni olish uchun 9n^{2} va -3n^{2} ni birlashtirish.

$$6n^{2}-23n+20=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 6 ni a, -23 ni b va 20 ni c bilan almashtiring.

$$n=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 6\times 20}}{2\times 6}$$

-23 kvadratini chiqarish.

$$n=\frac{-\left(-23\right)±\sqrt{529-4\times 6\times 20}}{2\times 6}$$

-4 ni 6 marotabaga ko'paytirish.

$$n=\frac{-\left(-23\right)±\sqrt{529-24\times 20}}{2\times 6}$$

-24 ni 20 marotabaga ko'paytirish.

$$n=\frac{-\left(-23\right)±\sqrt{529-480}}{2\times 6}$$

529 ni -480 ga qo'shish.

$$n=\frac{-\left(-23\right)±\sqrt{49}}{2\times 6}$$

49 ning kvadrat ildizini chiqarish.

$$n=\frac{-\left(-23\right)±7}{2\times 6}$$

-23 ning teskarisi 23 ga teng.

$$n=\frac{23±7}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$n=\frac{23±7}{12}$$

n=\frac{23±7}{12} tenglamasini yeching, bunda ± musbat. 23 ni 7 ga qo'shish.

$$n=\frac{30}{12}$$

\frac{30}{12} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$n=\frac{5}{2}$$

n=\frac{23±7}{12} tenglamasini yeching, bunda ± manfiy. 23 dan 7 ni ayirish.

$$n=\frac{16}{12}$$

\frac{16}{12} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$n=\frac{4}{3}$$

Tenglama yechildi.

$$n=\frac{5}{2} n=\frac{4}{3}$$

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan 3n^{2} ni ayirish.

$$9n^{2}-23n+20-3n^{2}=0$$

6n^{2} ni olish uchun 9n^{2} va -3n^{2} ni birlashtirish.

$$6n^{2}-23n+20=0$$

Ikkala tarafdan 20 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

$$6n^{2}-23n=-20$$

Ikki tarafini 6 ga bo‘ling.

$$\frac{6n^{2}-23n}{6}=-\frac{20}{6}$$

6 ga bo'lish 6 ga ko'paytirishni bekor qiladi.

$$n^{2}-\frac{23}{6}n=-\frac{20}{6}$$

\frac{-20}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$n^{2}-\frac{23}{6}n=-\frac{10}{3}$$

-\frac{23}{6} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{23}{12} olish uchun. Keyin, -\frac{23}{12} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$n^{2}-\frac{23}{6}n+\left(-\frac{23}{12}\right)^{2}=-\frac{10}{3}+\left(-\frac{23}{12}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{23}{12} kvadratini chiqarish.

$$n^{2}-\frac{23}{6}n+\frac{529}{144}=-\frac{10}{3}+\frac{529}{144}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{10}{3} ni \frac{529}{144} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$n^{2}-\frac{23}{6}n+\frac{529}{144}=\frac{49}{144}$$

n^{2}-\frac{23}{6}n+\frac{529}{144} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(n-\frac{23}{12}\right)^{2}=\frac{49}{144}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(n-\frac{23}{12}\right)^{2}}=\sqrt{\frac{49}{144}}$$

Qisqartirish.

$$n-\frac{23}{12}=\frac{7}{12} n-\frac{23}{12}=-\frac{7}{12}$$

\frac{23}{12} ni tenglamaning ikkala tarafiga qo'shish.

$$n=\frac{5}{2} n=\frac{4}{3}$$