11y=3y^2-4 eching

$$11 y = 3 y ^ { 2 } - 4$$

y uchun yechish

$$y=4$$

$$y=-\frac{1}{3}\approx -0,333333333$$

Guruhlash orqali faktorlash qadamlari

Ikkala tarafdan 3y^{2} ni ayirish.

$$11y-3y^{2}=-4$$

4 ni ikki tarafga qo’shing.

$$11y-3y^{2}+4=0$$

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

$$-3y^{2}+11y+4=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon -3y^{2}+ay+by+4 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=11 ab=-3\times 4=-12$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -12-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,12 -2,6 -3,4$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+12=11 -2+6=4 -3+4=1$$

Yechim – 11 yigʻindisini beruvchi juftlik.

$$a=12 b=-1$$

-3y^{2}+11y+4 ni \left(-3y^{2}+12y\right)+\left(-y+4\right) sifatida qaytadan yozish.

$$\left(-3y^{2}+12y\right)+\left(-y+4\right)$$

-3y^{2}+12y ichida 3y ni ajrating.

$$3y\left(-y+4\right)-y+4$$

Distributiv funktsiyasidan foydalangan holda -y+4 umumiy terminini chiqaring.

$$\left(-y+4\right)\left(3y+1\right)$$

Tenglamani yechish uchun -y+4=0 va 3y+1=0 ni yeching.

$$y=4 y=-\frac{1}{3}$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan 3y^{2} ni ayirish.

$$11y-3y^{2}=-4$$

4 ni ikki tarafga qo’shing.

$$11y-3y^{2}+4=0$$

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.

$$-3y^{2}+11y+4=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -3 ni a, 11 ni b va 4 ni c bilan almashtiring.

$$y=\frac{-11±\sqrt{11^{2}-4\left(-3\right)\times 4}}{2\left(-3\right)}$$

11 kvadratini chiqarish.

$$y=\frac{-11±\sqrt{121-4\left(-3\right)\times 4}}{2\left(-3\right)}$$

-4 ni -3 marotabaga ko'paytirish.

$$y=\frac{-11±\sqrt{121+12\times 4}}{2\left(-3\right)}$$

12 ni 4 marotabaga ko'paytirish.

$$y=\frac{-11±\sqrt{121+48}}{2\left(-3\right)}$$

121 ni 48 ga qo'shish.

$$y=\frac{-11±\sqrt{169}}{2\left(-3\right)}$$

169 ning kvadrat ildizini chiqarish.

$$y=\frac{-11±13}{2\left(-3\right)}$$

2 ni -3 marotabaga ko'paytirish.

$$y=\frac{-11±13}{-6}$$

y=\frac{-11±13}{-6} tenglamasini yeching, bunda ± musbat. -11 ni 13 ga qo'shish.

$$y=\frac{2}{-6}$$

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

$$y=-\frac{1}{3}$$

y=\frac{-11±13}{-6} tenglamasini yeching, bunda ± manfiy. -11 dan 13 ni ayirish.

$$y=-\frac{24}{-6}$$

-24 ni -6 ga bo'lish.

$$y=4$$

Tenglama yechildi.

$$y=-\frac{1}{3} y=4$$

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan 3y^{2} ni ayirish.

$$11y-3y^{2}=-4$$

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$-3y^{2}+11y=-4$$

Ikki tarafini -3 ga bo‘ling.

$$\frac{-3y^{2}+11y}{-3}=-\frac{4}{-3}$$

-3 ga bo'lish -3 ga ko'paytirishni bekor qiladi.

$$y^{2}+\frac{11}{-3}y=-\frac{4}{-3}$$

11 ni -3 ga bo'lish.

$$y^{2}-\frac{11}{3}y=-\frac{4}{-3}$$

-4 ni -3 ga bo'lish.

$$y^{2}-\frac{11}{3}y=\frac{4}{3}$$

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

$$y^{2}-\frac{11}{3}y+\left(-\frac{11}{6}\right)^{2}=\frac{4}{3}+\left(-\frac{11}{6}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{11}{6} kvadratini chiqarish.

$$y^{2}-\frac{11}{3}y+\frac{121}{36}=\frac{4}{3}+\frac{121}{36}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{4}{3} ni \frac{121}{36} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$y^{2}-\frac{11}{3}y+\frac{121}{36}=\frac{169}{36}$$

y^{2}-\frac{11}{3}y+\frac{121}{36} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(y-\frac{11}{6}\right)^{2}=\frac{169}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(y-\frac{11}{6}\right)^{2}}=\sqrt{\frac{169}{36}}$$

Qisqartirish.

$$y-\frac{11}{6}=\frac{13}{6} y-\frac{11}{6}=-\frac{13}{6}$$

\frac{11}{6} ni tenglamaning ikkala tarafiga qo'shish.

$$y=4 y=-\frac{1}{3}$$