8-26y+11y^2=0 eching

$$8 - 26 y + 11 y ^ { 2 } = 0$$

y uchun yechish

$$y=2$$

$$y=\frac{4}{11}\approx 0,363636364$$

Guruhlash orqali faktorlash qadamlari

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

$$11y^{2}-26y+8=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 11y^{2}+ay+by+8 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-26 ab=11\times 8=88$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 88-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-88 -2,-44 -4,-22 -8,-11$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-88=-89 -2-44=-46 -4-22=-26 -8-11=-19$$

Yechim – -26 yigʻindisini beruvchi juftlik.

$$a=-22 b=-4$$

11y^{2}-26y+8 ni \left(11y^{2}-22y\right)+\left(-4y+8\right) sifatida qaytadan yozish.

$$\left(11y^{2}-22y\right)+\left(-4y+8\right)$$

Birinchi guruhda 11y ni va ikkinchi guruhda -4 ni faktordan chiqaring.

$$11y\left(y-2\right)-4\left(y-2\right)$$

Distributiv funktsiyasidan foydalangan holda y-2 umumiy terminini chiqaring.

$$\left(y-2\right)\left(11y-4\right)$$

Tenglamani yechish uchun y-2=0 va 11y-4=0 ni yeching.

$$y=2 y=\frac{4}{11}$$

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.

$$11y^{2}-26y+8=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 11 ni a, -26 ni b va 8 ni c bilan almashtiring.

$$y=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 11\times 8}}{2\times 11}$$

-26 kvadratini chiqarish.

$$y=\frac{-\left(-26\right)±\sqrt{676-4\times 11\times 8}}{2\times 11}$$

-4 ni 11 marotabaga ko'paytirish.

$$y=\frac{-\left(-26\right)±\sqrt{676-44\times 8}}{2\times 11}$$

-44 ni 8 marotabaga ko'paytirish.

$$y=\frac{-\left(-26\right)±\sqrt{676-352}}{2\times 11}$$

676 ni -352 ga qo'shish.

$$y=\frac{-\left(-26\right)±\sqrt{324}}{2\times 11}$$

324 ning kvadrat ildizini chiqarish.

$$y=\frac{-\left(-26\right)±18}{2\times 11}$$

-26 ning teskarisi 26 ga teng.

$$y=\frac{26±18}{2\times 11}$$

2 ni 11 marotabaga ko'paytirish.

$$y=\frac{26±18}{22}$$

y=\frac{26±18}{22} tenglamasini yeching, bunda ± musbat. 26 ni 18 ga qo'shish.

$$y=\frac{44}{22}$$

44 ni 22 ga bo'lish.

$$y=2$$

y=\frac{26±18}{22} tenglamasini yeching, bunda ± manfiy. 26 dan 18 ni ayirish.

$$y=\frac{8}{22}$$

\frac{8}{22} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$y=\frac{4}{11}$$

Tenglama yechildi.

$$y=2 y=\frac{4}{11}$$

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.

$$11y^{2}-26y+8=0$$

Tenglamaning ikkala tarafidan 8 ni ayirish.

$$11y^{2}-26y+8-8=-8$$

O‘zidan 8 ayirilsa 0 qoladi.

$$11y^{2}-26y=-8$$

Ikki tarafini 11 ga bo‘ling.

$$\frac{11y^{2}-26y}{11}=-\frac{8}{11}$$

11 ga bo'lish 11 ga ko'paytirishni bekor qiladi.

$$y^{2}-\frac{26}{11}y=-\frac{8}{11}$$

-\frac{26}{11} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{13}{11} olish uchun. Keyin, -\frac{13}{11} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$y^{2}-\frac{26}{11}y+\left(-\frac{13}{11}\right)^{2}=-\frac{8}{11}+\left(-\frac{13}{11}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{13}{11} kvadratini chiqarish.

$$y^{2}-\frac{26}{11}y+\frac{169}{121}=-\frac{8}{11}+\frac{169}{121}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{8}{11} ni \frac{169}{121} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$y^{2}-\frac{26}{11}y+\frac{169}{121}=\frac{81}{121}$$

y^{2}-\frac{26}{11}y+\frac{169}{121} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(y-\frac{13}{11}\right)^{2}=\frac{81}{121}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(y-\frac{13}{11}\right)^{2}}=\sqrt{\frac{81}{121}}$$

Qisqartirish.

$$y-\frac{13}{11}=\frac{9}{11} y-\frac{13}{11}=-\frac{9}{11}$$

\frac{13}{11} ni tenglamaning ikkala tarafiga qo'shish.

$$y=2 y=\frac{4}{11}$$