44j^2+24j+1=0 eching

$$44 j ^ { 2 } + 24 j + 1 = 0$$

j uchun yechish

$$j=-\frac{1}{22}\approx -0,045454545$$

$$j=-\frac{1}{2}=-0,5$$

Guruhlash orqali faktorlash qadamlari

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 44j^{2}+aj+bj+1 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=24 ab=44\times 1=44$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b musbat boʻlganda, a va b ikkisi ham musbat. 44-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,44 2,22 4,11$$

Har bir juftlik yigʻindisini hisoblang.

$$1+44=45 2+22=24 4+11=15$$

Yechim – 24 yigʻindisini beruvchi juftlik.

$$a=2 b=22$$

44j^{2}+24j+1 ni \left(44j^{2}+2j\right)+\left(22j+1\right) sifatida qaytadan yozish.

$$\left(44j^{2}+2j\right)+\left(22j+1\right)$$

44j^{2}+2j ichida 2j ni ajrating.

$$2j\left(22j+1\right)+22j+1$$

Distributiv funktsiyasidan foydalangan holda 22j+1 umumiy terminini chiqaring.

$$\left(22j+1\right)\left(2j+1\right)$$

Tenglamani yechish uchun 22j+1=0 va 2j+1=0 ni yeching.

$$j=-\frac{1}{22} j=-\frac{1}{2}$$

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.

$$44j^{2}+24j+1=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 44 ni a, 24 ni b va 1 ni c bilan almashtiring.

$$j=\frac{-24±\sqrt{24^{2}-4\times 44}}{2\times 44}$$

24 kvadratini chiqarish.

$$j=\frac{-24±\sqrt{576-4\times 44}}{2\times 44}$$

-4 ni 44 marotabaga ko'paytirish.

$$j=\frac{-24±\sqrt{576-176}}{2\times 44}$$

576 ni -176 ga qo'shish.

$$j=\frac{-24±\sqrt{400}}{2\times 44}$$

400 ning kvadrat ildizini chiqarish.

$$j=\frac{-24±20}{2\times 44}$$

2 ni 44 marotabaga ko'paytirish.

$$j=\frac{-24±20}{88}$$

j=\frac{-24±20}{88} tenglamasini yeching, bunda ± musbat. -24 ni 20 ga qo'shish.

$$j=-\frac{4}{88}$$

\frac{-4}{88} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$j=-\frac{1}{22}$$

j=\frac{-24±20}{88} tenglamasini yeching, bunda ± manfiy. -24 dan 20 ni ayirish.

$$j=-\frac{44}{88}$$

\frac{-44}{88} ulushini 44 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$j=-\frac{1}{2}$$

Tenglama yechildi.

$$j=-\frac{1}{22} j=-\frac{1}{2}$$

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.

$$44j^{2}+24j+1=0$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$44j^{2}+24j+1-1=-1$$

O‘zidan 1 ayirilsa 0 qoladi.

$$44j^{2}+24j=-1$$

Ikki tarafini 44 ga bo‘ling.

$$\frac{44j^{2}+24j}{44}=-\frac{1}{44}$$

44 ga bo'lish 44 ga ko'paytirishni bekor qiladi.

$$j^{2}+\frac{24}{44}j=-\frac{1}{44}$$

\frac{24}{44} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$j^{2}+\frac{6}{11}j=-\frac{1}{44}$$

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

$$j^{2}+\frac{6}{11}j+\left(\frac{3}{11}\right)^{2}=-\frac{1}{44}+\left(\frac{3}{11}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{11} kvadratini chiqarish.

$$j^{2}+\frac{6}{11}j+\frac{9}{121}=-\frac{1}{44}+\frac{9}{121}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{1}{44} ni \frac{9}{121} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$j^{2}+\frac{6}{11}j+\frac{9}{121}=\frac{25}{484}$$

j^{2}+\frac{6}{11}j+\frac{9}{121} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(j+\frac{3}{11}\right)^{2}=\frac{25}{484}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(j+\frac{3}{11}\right)^{2}}=\sqrt{\frac{25}{484}}$$

Qisqartirish.

$$j+\frac{3}{11}=\frac{5}{22} j+\frac{3}{11}=-\frac{5}{22}$$

Tenglamaning ikkala tarafidan \frac{3}{11} ni ayirish.

$$j=-\frac{1}{22} j=-\frac{1}{2}$$