10y^2-19y+6=0 eching

$$10 { y }^{ 2 } -19y+6 = 0$$

y uchun yechish

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

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

Guruhlash orqali faktorlash qadamlari

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 10y^{2}+ay+by+6 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-19 ab=10\times 6=60$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 60-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

Yechim – -19 yigʻindisini beruvchi juftlik.

$$a=-15 b=-4$$

10y^{2}-19y+6 ni \left(10y^{2}-15y\right)+\left(-4y+6\right) sifatida qaytadan yozish.

$$\left(10y^{2}-15y\right)+\left(-4y+6\right)$$

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

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

Distributiv funktsiyasidan foydalangan holda 2y-3 umumiy terminini chiqaring.

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

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

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

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.

$$10y^{2}-19y+6=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 10 ni a, -19 ni b va 6 ni c bilan almashtiring.

$$y=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 10\times 6}}{2\times 10}$$

-19 kvadratini chiqarish.

$$y=\frac{-\left(-19\right)±\sqrt{361-4\times 10\times 6}}{2\times 10}$$

-4 ni 10 marotabaga ko'paytirish.

$$y=\frac{-\left(-19\right)±\sqrt{361-40\times 6}}{2\times 10}$$

-40 ni 6 marotabaga ko'paytirish.

$$y=\frac{-\left(-19\right)±\sqrt{361-240}}{2\times 10}$$

361 ni -240 ga qo'shish.

$$y=\frac{-\left(-19\right)±\sqrt{121}}{2\times 10}$$

121 ning kvadrat ildizini chiqarish.

$$y=\frac{-\left(-19\right)±11}{2\times 10}$$

-19 ning teskarisi 19 ga teng.

$$y=\frac{19±11}{2\times 10}$$

2 ni 10 marotabaga ko'paytirish.

$$y=\frac{19±11}{20}$$

y=\frac{19±11}{20} tenglamasini yeching, bunda ± musbat. 19 ni 11 ga qo'shish.

$$y=\frac{30}{20}$$

\frac{30}{20} ulushini 10 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

y=\frac{19±11}{20} tenglamasini yeching, bunda ± manfiy. 19 dan 11 ni ayirish.

$$y=\frac{8}{20}$$

\frac{8}{20} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

Tenglama yechildi.

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

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.

$$10y^{2}-19y+6=0$$

Tenglamaning ikkala tarafidan 6 ni ayirish.

$$10y^{2}-19y+6-6=-6$$

O‘zidan 6 ayirilsa 0 qoladi.

$$10y^{2}-19y=-6$$

Ikki tarafini 10 ga bo‘ling.

$$\frac{10y^{2}-19y}{10}=-\frac{6}{10}$$

10 ga bo'lish 10 ga ko'paytirishni bekor qiladi.

$$y^{2}-\frac{19}{10}y=-\frac{6}{10}$$

\frac{-6}{10} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$y^{2}-\frac{19}{10}y=-\frac{3}{5}$$

-\frac{19}{10} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{19}{20} olish uchun. Keyin, -\frac{19}{20} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$y^{2}-\frac{19}{10}y+\left(-\frac{19}{20}\right)^{2}=-\frac{3}{5}+\left(-\frac{19}{20}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{19}{20} kvadratini chiqarish.

$$y^{2}-\frac{19}{10}y+\frac{361}{400}=-\frac{3}{5}+\frac{361}{400}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{3}{5} ni \frac{361}{400} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$y^{2}-\frac{19}{10}y+\frac{361}{400}=\frac{121}{400}$$

y^{2}-\frac{19}{10}y+\frac{361}{400} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(y-\frac{19}{20}\right)^{2}=\frac{121}{400}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(y-\frac{19}{20}\right)^{2}}=\sqrt{\frac{121}{400}}$$

Qisqartirish.

$$y-\frac{19}{20}=\frac{11}{20} y-\frac{19}{20}=-\frac{11}{20}$$

\frac{19}{20} ni tenglamaning ikkala tarafiga qo'shish.

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