4y^2-4y-8=0 eching

$$4 { y }^{ 2 } -4y-8=0$$

y uchun yechish

$$y=-1$$

$$y=2$$

Guruhlash orqali faktorlash qadamlari

Ikki tarafini 4 ga bo‘ling.

$$y^{2}-y-2=0$$

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

$$a+b=-1 ab=1\left(-2\right)=-2$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. Faqat bundan juftlik tizim yechimidir.

$$a=-2 b=1$$

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

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

y^{2}-2y ichida y ni ajrating.

$$y\left(y-2\right)+y-2$$

Distributiv funktsiyasidan foydalangan holda y-2 umumiy terminini chiqaring.

$$\left(y-2\right)\left(y+1\right)$$

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

$$y=2 y=-1$$

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.

$$4y^{2}-4y-8=0$$

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

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

-4 kvadratini chiqarish.

$$y=\frac{-\left(-4\right)±\sqrt{16-4\times 4\left(-8\right)}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$y=\frac{-\left(-4\right)±\sqrt{16-16\left(-8\right)}}{2\times 4}$$

-16 ni -8 marotabaga ko'paytirish.

$$y=\frac{-\left(-4\right)±\sqrt{16+128}}{2\times 4}$$

16 ni 128 ga qo'shish.

$$y=\frac{-\left(-4\right)±\sqrt{144}}{2\times 4}$$

144 ning kvadrat ildizini chiqarish.

$$y=\frac{-\left(-4\right)±12}{2\times 4}$$

-4 ning teskarisi 4 ga teng.

$$y=\frac{4±12}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$y=\frac{4±12}{8}$$

y=\frac{4±12}{8} tenglamasini yeching, bunda ± musbat. 4 ni 12 ga qo'shish.

$$y=\frac{16}{8}$$

16 ni 8 ga bo'lish.

$$y=2$$

y=\frac{4±12}{8} tenglamasini yeching, bunda ± manfiy. 4 dan 12 ni ayirish.

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

-8 ni 8 ga bo'lish.

$$y=-1$$

Tenglama yechildi.

$$y=2 y=-1$$

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.

$$4y^{2}-4y-8=0$$

8 ni tenglamaning ikkala tarafiga qo'shish.

$$4y^{2}-4y-8-\left(-8\right)=-\left(-8\right)$$

O‘zidan -8 ayirilsa 0 qoladi.

$$4y^{2}-4y=-\left(-8\right)$$

0 dan -8 ni ayirish.

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

Ikki tarafini 4 ga bo‘ling.

$$\frac{4y^{2}-4y}{4}=\frac{8}{4}$$

4 ga bo'lish 4 ga ko'paytirishni bekor qiladi.

$$y^{2}+\left(-\frac{4}{4}\right)y=\frac{8}{4}$$

-4 ni 4 ga bo'lish.

$$y^{2}-y=\frac{8}{4}$$

8 ni 4 ga bo'lish.

$$y^{2}-y=2$$

-1 ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{2} olish uchun. Keyin, -\frac{1}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$y^{2}-y+\left(-\frac{1}{2}\right)^{2}=2+\left(-\frac{1}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{2} kvadratini chiqarish.

$$y^{2}-y+\frac{1}{4}=2+\frac{1}{4}$$

2 ni \frac{1}{4} ga qo'shish.

$$y^{2}-y+\frac{1}{4}=\frac{9}{4}$$

y^{2}-y+\frac{1}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(y-\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(y-\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Qisqartirish.

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

\frac{1}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$y=2 y=-1$$