0=5n^2+1205n-90300 eching

$$0 = 5 n ^ { 2 } + 1205 n - 90300$$

n uchun yechish

$$n=-301$$

$$n=60$$

Guruhlash orqali faktorlash qadamlari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$5n^{2}+1205n-90300=0$$

Ikki tarafini 5 ga bo‘ling.

$$n^{2}+241n-18060=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon n^{2}+an+bn-18060 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=241 ab=1\left(-18060\right)=-18060$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -18060-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,18060 -2,9030 -3,6020 -4,4515 -5,3612 -6,3010 -7,2580 -10,1806 -12,1505 -14,1290 -15,1204 -20,903 -21,860 -28,645 -30,602 -35,516 -42,430 -43,420 -60,301 -70,258 -84,215 -86,210 -105,172 -129,140$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+18060=18059 -2+9030=9028 -3+6020=6017 -4+4515=4511 -5+3612=3607 -6+3010=3004 -7+2580=2573 -10+1806=1796 -12+1505=1493 -14+1290=1276 -15+1204=1189 -20+903=883 -21+860=839 -28+645=617 -30+602=572 -35+516=481 -42+430=388 -43+420=377 -60+301=241 -70+258=188 -84+215=131 -86+210=124 -105+172=67 -129+140=11$$

Yechim – 241 yigʻindisini beruvchi juftlik.

$$a=-60 b=301$$

n^{2}+241n-18060 ni \left(n^{2}-60n\right)+\left(301n-18060\right) sifatida qaytadan yozish.

$$\left(n^{2}-60n\right)+\left(301n-18060\right)$$

Birinchi guruhda n ni va ikkinchi guruhda 301 ni faktordan chiqaring.

$$n\left(n-60\right)+301\left(n-60\right)$$

Distributiv funktsiyasidan foydalangan holda n-60 umumiy terminini chiqaring.

$$\left(n-60\right)\left(n+301\right)$$

Tenglamani yechish uchun n-60=0 va n+301=0 ni yeching.

$$n=60 n=-301$$

Kvadrat tenglamalaridan foydalanish qoidalari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$5n^{2}+1205n-90300=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 1205 ni b va -90300 ni c bilan almashtiring.

$$n=\frac{-1205±\sqrt{1205^{2}-4\times 5\left(-90300\right)}}{2\times 5}$$

1205 kvadratini chiqarish.

$$n=\frac{-1205±\sqrt{1452025-4\times 5\left(-90300\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$n=\frac{-1205±\sqrt{1452025-20\left(-90300\right)}}{2\times 5}$$

-20 ni -90300 marotabaga ko'paytirish.

$$n=\frac{-1205±\sqrt{1452025+1806000}}{2\times 5}$$

1452025 ni 1806000 ga qo'shish.

$$n=\frac{-1205±\sqrt{3258025}}{2\times 5}$$

3258025 ning kvadrat ildizini chiqarish.

$$n=\frac{-1205±1805}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$n=\frac{-1205±1805}{10}$$

n=\frac{-1205±1805}{10} tenglamasini yeching, bunda ± musbat. -1205 ni 1805 ga qo'shish.

$$n=\frac{600}{10}$$

600 ni 10 ga bo'lish.

$$n=60$$

n=\frac{-1205±1805}{10} tenglamasini yeching, bunda ± manfiy. -1205 dan 1805 ni ayirish.

$$n=-\frac{3010}{10}$$

-3010 ni 10 ga bo'lish.

$$n=-301$$

Tenglama yechildi.

$$n=60 n=-301$$

Kvadratni yakunlash bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$5n^{2}+1205n-90300=0$$

90300 ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

$$5n^{2}+1205n=90300$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5n^{2}+1205n}{5}=\frac{90300}{5}$$

5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.

$$n^{2}+\frac{1205}{5}n=\frac{90300}{5}$$

1205 ni 5 ga bo'lish.

$$n^{2}+241n=\frac{90300}{5}$$

90300 ni 5 ga bo'lish.

$$n^{2}+241n=18060$$

241 ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{241}{2} olish uchun. Keyin, \frac{241}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$n^{2}+241n+\left(\frac{241}{2}\right)^{2}=18060+\left(\frac{241}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{241}{2} kvadratini chiqarish.

$$n^{2}+241n+\frac{58081}{4}=18060+\frac{58081}{4}$$

18060 ni \frac{58081}{4} ga qo'shish.

$$n^{2}+241n+\frac{58081}{4}=\frac{130321}{4}$$

n^{2}+241n+\frac{58081}{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.

$$\left(n+\frac{241}{2}\right)^{2}=\frac{130321}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(n+\frac{241}{2}\right)^{2}}=\sqrt{\frac{130321}{4}}$$

Qisqartirish.

$$n+\frac{241}{2}=\frac{361}{2} n+\frac{241}{2}=-\frac{361}{2}$$

Tenglamaning ikkala tarafidan \frac{241}{2} ni ayirish.

$$n=60 n=-301$$