2x^2-2x-12=28 eching

$$2 x ^ { 2 } - 2 x - 12 = 28$$

x uchun yechish

$$x=-4$$

$$x=5$$

Guruhlash orqali faktorlash qadamlari

Ikkala tarafdan 28 ni ayirish.

$$2x^{2}-2x-12-28=0$$

-40 olish uchun -12 dan 28 ni ayirish.

$$2x^{2}-2x-40=0$$

Ikki tarafini 2 ga bo‘ling.

$$x^{2}-x-20=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon x^{2}+ax+bx-20 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-1 ab=1\left(-20\right)=-20$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -20-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-20 2,-10 4,-5$$

Har bir juftlik yigʻindisini hisoblang.

$$1-20=-19 2-10=-8 4-5=-1$$

Yechim – -1 yigʻindisini beruvchi juftlik.

$$a=-5 b=4$$

x^{2}-x-20 ni \left(x^{2}-5x\right)+\left(4x-20\right) sifatida qaytadan yozish.

$$\left(x^{2}-5x\right)+\left(4x-20\right)$$

Birinchi guruhda x ni va ikkinchi guruhda 4 ni faktordan chiqaring.

$$x\left(x-5\right)+4\left(x-5\right)$$

Distributiv funktsiyasidan foydalangan holda x-5 umumiy terminini chiqaring.

$$\left(x-5\right)\left(x+4\right)$$

Tenglamani yechish uchun x-5=0 va x+4=0 ni yeching.

$$x=5 x=-4$$

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.

$$2x^{2}-2x-12=28$$

Tenglamaning ikkala tarafidan 28 ni ayirish.

$$2x^{2}-2x-12-28=28-28$$

O‘zidan 28 ayirilsa 0 qoladi.

$$2x^{2}-2x-12-28=0$$

-12 dan 28 ni ayirish.

$$2x^{2}-2x-40=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, -2 ni b va -40 ni c bilan almashtiring.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 2\left(-40\right)}}{2\times 2}$$

-2 kvadratini chiqarish.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 2\left(-40\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4-8\left(-40\right)}}{2\times 2}$$

-8 ni -40 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4+320}}{2\times 2}$$

4 ni 320 ga qo'shish.

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

324 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-2\right)±18}{2\times 2}$$

-2 ning teskarisi 2 ga teng.

$$x=\frac{2±18}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{2±18}{4}$$

x=\frac{2±18}{4} tenglamasini yeching, bunda ± musbat. 2 ni 18 ga qo'shish.

$$x=\frac{20}{4}$$

20 ni 4 ga bo'lish.

$$x=5$$

x=\frac{2±18}{4} tenglamasini yeching, bunda ± manfiy. 2 dan 18 ni ayirish.

$$x=-\frac{16}{4}$$

-16 ni 4 ga bo'lish.

$$x=-4$$

Tenglama yechildi.

$$x=5 x=-4$$

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.

$$2x^{2}-2x-12=28$$

12 ni tenglamaning ikkala tarafiga qo'shish.

$$2x^{2}-2x-12-\left(-12\right)=28-\left(-12\right)$$

O‘zidan -12 ayirilsa 0 qoladi.

$$2x^{2}-2x=28-\left(-12\right)$$

28 dan -12 ni ayirish.

$$2x^{2}-2x=40$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}-2x}{2}=\frac{40}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{2}{2}\right)x=\frac{40}{2}$$

-2 ni 2 ga bo'lish.

$$x^{2}-x=\frac{40}{2}$$

40 ni 2 ga bo'lish.

$$x^{2}-x=20$$

-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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=20+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=20+\frac{1}{4}$$

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

$$x^{2}-x+\frac{1}{4}=\frac{81}{4}$$

x^{2}-x+\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.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{81}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Qisqartirish.

$$x-\frac{1}{2}=\frac{9}{2} x-\frac{1}{2}=-\frac{9}{2}$$

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

$$x=5 x=-4$$