6y^2-17y+5=0 eching

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

y uchun yechish

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

$$y = \frac{5}{2} = 2\frac{1}{2} = 2,5$$

Guruhlash orqali faktorlash qadamlari

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

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

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.

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

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

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

Birinchi guruhda 3y ni va ikkinchi guruhda -1 ni faktordan chiqaring.

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

Distributiv funktsiyasidan foydalangan holda 2y-5 umumiy terminini chiqaring.

$$\left(2y-5\right)\left(3y-1\right)$$

Tenglamani yechish uchun 2y-5=0 va 3y-1=0 ni yeching.

$$y=\frac{5}{2} y=\frac{1}{3}$$

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.

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

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

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

-17 kvadratini chiqarish.

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

-4 ni 6 marotabaga ko'paytirish.

$$y=\frac{-\left(-17\right)±\sqrt{289-24\times 5}}{2\times 6}$$

-24 ni 5 marotabaga ko'paytirish.

$$y=\frac{-\left(-17\right)±\sqrt{289-120}}{2\times 6}$$

289 ni -120 ga qo'shish.

$$y=\frac{-\left(-17\right)±\sqrt{169}}{2\times 6}$$

169 ning kvadrat ildizini chiqarish.

$$y=\frac{-\left(-17\right)±13}{2\times 6}$$

-17 ning teskarisi 17 ga teng.

$$y=\frac{17±13}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$y=\frac{17±13}{12}$$

y=\frac{17±13}{12} tenglamasini yeching, bunda ± musbat. 17 ni 13 ga qo'shish.

$$y=\frac{30}{12}$$

\frac{30}{12} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

y=\frac{17±13}{12} tenglamasini yeching, bunda ± manfiy. 17 dan 13 ni ayirish.

$$y=\frac{4}{12}$$

\frac{4}{12} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

Tenglama yechildi.

$$y=\frac{5}{2} y=\frac{1}{3}$$

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.

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

Tenglamaning ikkala tarafidan 5 ni ayirish.

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

O‘zidan 5 ayirilsa 0 qoladi.

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

Ikki tarafini 6 ga bo‘ling.

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

6 ga bo'lish 6 ga ko'paytirishni bekor qiladi.

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

-\frac{17}{6} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{17}{12} olish uchun. Keyin, -\frac{17}{12} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$y^{2}-\frac{17}{6}y+\left(-\frac{17}{12}\right)^{2}=-\frac{5}{6}+\left(-\frac{17}{12}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{17}{12} kvadratini chiqarish.

$$y^{2}-\frac{17}{6}y+\frac{289}{144}=-\frac{5}{6}+\frac{289}{144}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{5}{6} ni \frac{289}{144} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$y^{2}-\frac{17}{6}y+\frac{289}{144}=\frac{169}{144}$$

y^{2}-\frac{17}{6}y+\frac{289}{144} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(y-\frac{17}{12}\right)^{2}=\frac{169}{144}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(y-\frac{17}{12}\right)^{2}}=\sqrt{\frac{169}{144}}$$

Qisqartirish.

$$y-\frac{17}{12}=\frac{13}{12} y-\frac{17}{12}=-\frac{13}{12}$$

\frac{17}{12} ni tenglamaning ikkala tarafiga qo'shish.

$$y=\frac{5}{2} y=\frac{1}{3}$$