2b^2-105=-11b eching

$$2 b ^ { 2 } - 105 = - 11 b$$

b uchun yechish

$$b = -\frac{21}{2} = -10\frac{1}{2} = -10,5$$

$$b=5$$

Guruhlash orqali faktorlash qadamlari

11b ni ikki tarafga qo’shing.

$$2b^{2}-105+11b=0$$

Polinomni standart shaklga keltirish uchun uni qayta tartiblang. Shartlarni eng yuqoridan eng pastki qiymat ko'rsatgichiga joylashtirish.

$$2b^{2}+11b-105=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 2b^{2}+ab+bb-105 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=11 ab=2\left(-105\right)=-210$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -210-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,210 -2,105 -3,70 -5,42 -6,35 -7,30 -10,21 -14,15$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+210=209 -2+105=103 -3+70=67 -5+42=37 -6+35=29 -7+30=23 -10+21=11 -14+15=1$$

Yechim – 11 yigʻindisini beruvchi juftlik.

$$a=-10 b=21$$

2b^{2}+11b-105 ni \left(2b^{2}-10b\right)+\left(21b-105\right) sifatida qaytadan yozish.

$$\left(2b^{2}-10b\right)+\left(21b-105\right)$$

Birinchi guruhda 2b ni va ikkinchi guruhda 21 ni faktordan chiqaring.

$$2b\left(b-5\right)+21\left(b-5\right)$$

Distributiv funktsiyasidan foydalangan holda b-5 umumiy terminini chiqaring.

$$\left(b-5\right)\left(2b+21\right)$$

Tenglamani yechish uchun b-5=0 va 2b+21=0 ni yeching.

$$b=5 b=-\frac{21}{2}$$

Kvadrat tenglamalaridan foydalanish qoidalari

11b ni ikki tarafga qo’shing.

$$2b^{2}-105+11b=0$$

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.

$$2b^{2}+11b-105=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, 11 ni b va -105 ni c bilan almashtiring.

$$b=\frac{-11±\sqrt{11^{2}-4\times 2\left(-105\right)}}{2\times 2}$$

11 kvadratini chiqarish.

$$b=\frac{-11±\sqrt{121-4\times 2\left(-105\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$b=\frac{-11±\sqrt{121-8\left(-105\right)}}{2\times 2}$$

-8 ni -105 marotabaga ko'paytirish.

$$b=\frac{-11±\sqrt{121+840}}{2\times 2}$$

121 ni 840 ga qo'shish.

$$b=\frac{-11±\sqrt{961}}{2\times 2}$$

961 ning kvadrat ildizini chiqarish.

$$b=\frac{-11±31}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$b=\frac{-11±31}{4}$$

b=\frac{-11±31}{4} tenglamasini yeching, bunda ± musbat. -11 ni 31 ga qo'shish.

$$b=\frac{20}{4}$$

20 ni 4 ga bo'lish.

$$b=5$$

b=\frac{-11±31}{4} tenglamasini yeching, bunda ± manfiy. -11 dan 31 ni ayirish.

$$b=-\frac{42}{4}$$

\frac{-42}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$b=-\frac{21}{2}$$

Tenglama yechildi.

$$b=5 b=-\frac{21}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

11b ni ikki tarafga qo’shing.

$$2b^{2}-105+11b=0$$

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

$$2b^{2}+11b=105$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2b^{2}+11b}{2}=\frac{105}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$b^{2}+\frac{11}{2}b=\frac{105}{2}$$

\frac{11}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{11}{4} olish uchun. Keyin, \frac{11}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$b^{2}+\frac{11}{2}b+\left(\frac{11}{4}\right)^{2}=\frac{105}{2}+\left(\frac{11}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{11}{4} kvadratini chiqarish.

$$b^{2}+\frac{11}{2}b+\frac{121}{16}=\frac{105}{2}+\frac{121}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{105}{2} ni \frac{121}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$b^{2}+\frac{11}{2}b+\frac{121}{16}=\frac{961}{16}$$

b^{2}+\frac{11}{2}b+\frac{121}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(b+\frac{11}{4}\right)^{2}=\frac{961}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(b+\frac{11}{4}\right)^{2}}=\sqrt{\frac{961}{16}}$$

Qisqartirish.

$$b+\frac{11}{4}=\frac{31}{4} b+\frac{11}{4}=-\frac{31}{4}$$

Tenglamaning ikkala tarafidan \frac{11}{4} ni ayirish.

$$b=5 b=-\frac{21}{2}$$