5x^2-x-2184=0 eching

$$5 { x }^{ 2 } -x-2184=0$$

x uchun yechish

$$x = -\frac{104}{5} = -20\frac{4}{5} = -20,8$$

$$x=21$$

Guruhlash orqali faktorlash qadamlari

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 5x^{2}+ax+bx-2184 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-1 ab=5\left(-2184\right)=-10920$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -10920-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-10920 2,-5460 3,-3640 4,-2730 5,-2184 6,-1820 7,-1560 8,-1365 10,-1092 12,-910 13,-840 14,-780 15,-728 20,-546 21,-520 24,-455 26,-420 28,-390 30,-364 35,-312 39,-280 40,-273 42,-260 52,-210 56,-195 60,-182 65,-168 70,-156 78,-140 84,-130 91,-120 104,-105$$

Har bir juftlik yigʻindisini hisoblang.

$$1-10920=-10919 2-5460=-5458 3-3640=-3637 4-2730=-2726 5-2184=-2179 6-1820=-1814 7-1560=-1553 8-1365=-1357 10-1092=-1082 12-910=-898 13-840=-827 14-780=-766 15-728=-713 20-546=-526 21-520=-499 24-455=-431 26-420=-394 28-390=-362 30-364=-334 35-312=-277 39-280=-241 40-273=-233 42-260=-218 52-210=-158 56-195=-139 60-182=-122 65-168=-103 70-156=-86 78-140=-62 84-130=-46 91-120=-29 104-105=-1$$

Yechim – -1 yigʻindisini beruvchi juftlik.

$$a=-105 b=104$$

5x^{2}-x-2184 ni \left(5x^{2}-105x\right)+\left(104x-2184\right) sifatida qaytadan yozish.

$$\left(5x^{2}-105x\right)+\left(104x-2184\right)$$

Birinchi guruhda 5x ni va ikkinchi guruhda 104 ni faktordan chiqaring.

$$5x\left(x-21\right)+104\left(x-21\right)$$

Distributiv funktsiyasidan foydalangan holda x-21 umumiy terminini chiqaring.

$$\left(x-21\right)\left(5x+104\right)$$

Tenglamani yechish uchun x-21=0 va 5x+104=0 ni yeching.

$$x=21 x=-\frac{104}{5}$$

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.

$$5x^{2}-x-2184=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, -1 ni b va -2184 ni c bilan almashtiring.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 5\left(-2184\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\left(-1\right)±\sqrt{1-20\left(-2184\right)}}{2\times 5}$$

-20 ni -2184 marotabaga ko'paytirish.

$$x=\frac{-\left(-1\right)±\sqrt{1+43680}}{2\times 5}$$

1 ni 43680 ga qo'shish.

$$x=\frac{-\left(-1\right)±\sqrt{43681}}{2\times 5}$$

43681 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-1\right)±209}{2\times 5}$$

-1 ning teskarisi 1 ga teng.

$$x=\frac{1±209}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{1±209}{10}$$

x=\frac{1±209}{10} tenglamasini yeching, bunda ± musbat. 1 ni 209 ga qo'shish.

$$x=\frac{210}{10}$$

210 ni 10 ga bo'lish.

$$x=21$$

x=\frac{1±209}{10} tenglamasini yeching, bunda ± manfiy. 1 dan 209 ni ayirish.

$$x=-\frac{208}{10}$$

\frac{-208}{10} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{104}{5}$$

Tenglama yechildi.

$$x=21 x=-\frac{104}{5}$$

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.

$$5x^{2}-x-2184=0$$

2184 ni tenglamaning ikkala tarafiga qo'shish.

$$5x^{2}-x-2184-\left(-2184\right)=-\left(-2184\right)$$

O‘zidan -2184 ayirilsa 0 qoladi.

$$5x^{2}-x=-\left(-2184\right)$$

0 dan -2184 ni ayirish.

$$5x^{2}-x=2184$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}-x}{5}=\frac{2184}{5}$$

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

$$x^{2}-\frac{1}{5}x=\frac{2184}{5}$$

-\frac{1}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{10} olish uchun. Keyin, -\frac{1}{10} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=\frac{2184}{5}+\left(-\frac{1}{10}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{10} kvadratini chiqarish.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{2184}{5}+\frac{1}{100}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{2184}{5} ni \frac{1}{100} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{43681}{100}$$

x^{2}-\frac{1}{5}x+\frac{1}{100} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{10}\right)^{2}=\frac{43681}{100}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{10}\right)^{2}}=\sqrt{\frac{43681}{100}}$$

Qisqartirish.

$$x-\frac{1}{10}=\frac{209}{10} x-\frac{1}{10}=-\frac{209}{10}$$

\frac{1}{10} ni tenglamaning ikkala tarafiga qo'shish.

$$x=21 x=-\frac{104}{5}$$