3y(y-2)=4y-8 eching

$$3 y ( y - 2 ) = 4 y - 8$$

y uchun yechish

$$y=2$$

$$y = \frac{4}{3} = 1\frac{1}{3} \approx 1,333333333$$

Kvadrat tenglamalaridan foydalanish qoidalari

3y ga y-2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3y^{2}-6y=4y-8$$

Ikkala tarafdan 4y ni ayirish.

$$3y^{2}-6y-4y=-8$$

-10y ni olish uchun -6y va -4y ni birlashtirish.

$$3y^{2}-10y=-8$$

8 ni ikki tarafga qo’shing.

$$3y^{2}-10y+8=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, -10 ni b va 8 ni c bilan almashtiring.

$$y=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 3\times 8}}{2\times 3}$$

-10 kvadratini chiqarish.

$$y=\frac{-\left(-10\right)±\sqrt{100-4\times 3\times 8}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$y=\frac{-\left(-10\right)±\sqrt{100-12\times 8}}{2\times 3}$$

-12 ni 8 marotabaga ko'paytirish.

$$y=\frac{-\left(-10\right)±\sqrt{100-96}}{2\times 3}$$

100 ni -96 ga qo'shish.

$$y=\frac{-\left(-10\right)±\sqrt{4}}{2\times 3}$$

4 ning kvadrat ildizini chiqarish.

$$y=\frac{-\left(-10\right)±2}{2\times 3}$$

-10 ning teskarisi 10 ga teng.

$$y=\frac{10±2}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$y=\frac{10±2}{6}$$

y=\frac{10±2}{6} tenglamasini yeching, bunda ± musbat. 10 ni 2 ga qo'shish.

$$y=\frac{12}{6}$$

12 ni 6 ga bo'lish.

$$y=2$$

y=\frac{10±2}{6} tenglamasini yeching, bunda ± manfiy. 10 dan 2 ni ayirish.

$$y=\frac{8}{6}$$

\frac{8}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

Tenglama yechildi.

$$y=2 y=\frac{4}{3}$$

Kvadratni yakunlash bo'yicha qadamlar

3y ga y-2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3y^{2}-6y=4y-8$$

Ikkala tarafdan 4y ni ayirish.

$$3y^{2}-6y-4y=-8$$

-10y ni olish uchun -6y va -4y ni birlashtirish.

$$3y^{2}-10y=-8$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3y^{2}-10y}{3}=-\frac{8}{3}$$

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

$$y^{2}-\frac{10}{3}y=-\frac{8}{3}$$

-\frac{10}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{5}{3} olish uchun. Keyin, -\frac{5}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$y^{2}-\frac{10}{3}y+\left(-\frac{5}{3}\right)^{2}=-\frac{8}{3}+\left(-\frac{5}{3}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{5}{3} kvadratini chiqarish.

$$y^{2}-\frac{10}{3}y+\frac{25}{9}=-\frac{8}{3}+\frac{25}{9}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{8}{3} ni \frac{25}{9} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$y^{2}-\frac{10}{3}y+\frac{25}{9}=\frac{1}{9}$$

y^{2}-\frac{10}{3}y+\frac{25}{9} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(y-\frac{5}{3}\right)^{2}=\frac{1}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(y-\frac{5}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Qisqartirish.

$$y-\frac{5}{3}=\frac{1}{3} y-\frac{5}{3}=-\frac{1}{3}$$

\frac{5}{3} ni tenglamaning ikkala tarafiga qo'shish.

$$y=2 y=\frac{4}{3}$$