5y^2=17y-6 eching

$$5 y ^ { 2 } = 17 y - 6$$

y uchun yechish

$$y=\frac{2}{5}=0,4$$

$$y=3$$

Guruhlash orqali faktorlash qadamlari

Ikkala tarafdan 17y ni ayirish.

$$5y^{2}-17y=-6$$

6 ni ikki tarafga qo’shing.

$$5y^{2}-17y+6=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 5y^{2}+ay+by+6 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-17 ab=5\times 6=30$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 30-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

Yechim – -17 yigʻindisini beruvchi juftlik.

$$a=-15 b=-2$$

5y^{2}-17y+6 ni \left(5y^{2}-15y\right)+\left(-2y+6\right) sifatida qaytadan yozish.

$$\left(5y^{2}-15y\right)+\left(-2y+6\right)$$

Birinchi guruhda 5y ni va ikkinchi guruhda -2 ni faktordan chiqaring.

$$5y\left(y-3\right)-2\left(y-3\right)$$

Distributiv funktsiyasidan foydalangan holda y-3 umumiy terminini chiqaring.

$$\left(y-3\right)\left(5y-2\right)$$

Tenglamani yechish uchun y-3=0 va 5y-2=0 ni yeching.

$$y=3 y=\frac{2}{5}$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan 17y ni ayirish.

$$5y^{2}-17y=-6$$

6 ni ikki tarafga qo’shing.

$$5y^{2}-17y+6=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, -17 ni b va 6 ni c bilan almashtiring.

$$y=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 5\times 6}}{2\times 5}$$

-17 kvadratini chiqarish.

$$y=\frac{-\left(-17\right)±\sqrt{289-4\times 5\times 6}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$y=\frac{-\left(-17\right)±\sqrt{289-20\times 6}}{2\times 5}$$

-20 ni 6 marotabaga ko'paytirish.

$$y=\frac{-\left(-17\right)±\sqrt{289-120}}{2\times 5}$$

289 ni -120 ga qo'shish.

$$y=\frac{-\left(-17\right)±\sqrt{169}}{2\times 5}$$

169 ning kvadrat ildizini chiqarish.

$$y=\frac{-\left(-17\right)±13}{2\times 5}$$

-17 ning teskarisi 17 ga teng.

$$y=\frac{17±13}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$y=\frac{17±13}{10}$$

y=\frac{17±13}{10} tenglamasini yeching, bunda ± musbat. 17 ni 13 ga qo'shish.

$$y=\frac{30}{10}$$

30 ni 10 ga bo'lish.

$$y=3$$

y=\frac{17±13}{10} tenglamasini yeching, bunda ± manfiy. 17 dan 13 ni ayirish.

$$y=\frac{4}{10}$$

\frac{4}{10} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$y=\frac{2}{5}$$

Tenglama yechildi.

$$y=3 y=\frac{2}{5}$$

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan 17y ni ayirish.

$$5y^{2}-17y=-6$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5y^{2}-17y}{5}=-\frac{6}{5}$$

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

$$y^{2}-\frac{17}{5}y=-\frac{6}{5}$$

-\frac{17}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{17}{10} olish uchun. Keyin, -\frac{17}{10} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$y^{2}-\frac{17}{5}y+\left(-\frac{17}{10}\right)^{2}=-\frac{6}{5}+\left(-\frac{17}{10}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{17}{10} kvadratini chiqarish.

$$y^{2}-\frac{17}{5}y+\frac{289}{100}=-\frac{6}{5}+\frac{289}{100}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{6}{5} ni \frac{289}{100} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$y^{2}-\frac{17}{5}y+\frac{289}{100}=\frac{169}{100}$$

y^{2}-\frac{17}{5}y+\frac{289}{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.

$$\left(y-\frac{17}{10}\right)^{2}=\frac{169}{100}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(y-\frac{17}{10}\right)^{2}}=\sqrt{\frac{169}{100}}$$

Qisqartirish.

$$y-\frac{17}{10}=\frac{13}{10} y-\frac{17}{10}=-\frac{13}{10}$$

\frac{17}{10} ni tenglamaning ikkala tarafiga qo'shish.

$$y=3 y=\frac{2}{5}$$