-3x^2-5x+12=0 eching

$$- 3 x ^ { 2 } - 5 x + 12 = 0$$

x uchun yechish

$$x=-3$$

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1,333333333$$

Guruhlash orqali faktorlash qadamlari

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon -3x^{2}+ax+bx+12 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-5 ab=-3\times 12=-36$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -36-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Har bir juftlik yigʻindisini hisoblang.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

Yechim – -5 yigʻindisini beruvchi juftlik.

$$a=4 b=-9$$

-3x^{2}-5x+12 ni \left(-3x^{2}+4x\right)+\left(-9x+12\right) sifatida qaytadan yozish.

$$\left(-3x^{2}+4x\right)+\left(-9x+12\right)$$

Birinchi guruhda -x ni va ikkinchi guruhda -3 ni faktordan chiqaring.

$$-x\left(3x-4\right)-3\left(3x-4\right)$$

Distributiv funktsiyasidan foydalangan holda 3x-4 umumiy terminini chiqaring.

$$\left(3x-4\right)\left(-x-3\right)$$

Tenglamani yechish uchun 3x-4=0 va -x-3=0 ni yeching.

$$x=\frac{4}{3} x=-3$$

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.

$$-3x^{2}-5x+12=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -3 ni a, -5 ni b va 12 ni c bilan almashtiring.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-3\right)\times 12}}{2\left(-3\right)}$$

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-3\right)\times 12}}{2\left(-3\right)}$$

-4 ni -3 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+12\times 12}}{2\left(-3\right)}$$

12 ni 12 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+144}}{2\left(-3\right)}$$

25 ni 144 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{169}}{2\left(-3\right)}$$

169 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-5\right)±13}{2\left(-3\right)}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±13}{2\left(-3\right)}$$

2 ni -3 marotabaga ko'paytirish.

$$x=\frac{5±13}{-6}$$

x=\frac{5±13}{-6} tenglamasini yeching, bunda ± musbat. 5 ni 13 ga qo'shish.

$$x=\frac{18}{-6}$$

18 ni -6 ga bo'lish.

$$x=-3$$

x=\frac{5±13}{-6} tenglamasini yeching, bunda ± manfiy. 5 dan 13 ni ayirish.

$$x=-\frac{8}{-6}$$

\frac{-8}{-6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{4}{3}$$

Tenglama yechildi.

$$x=-3 x=\frac{4}{3}$$

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.

$$-3x^{2}-5x+12=0$$

Tenglamaning ikkala tarafidan 12 ni ayirish.

$$-3x^{2}-5x+12-12=-12$$

O‘zidan 12 ayirilsa 0 qoladi.

$$-3x^{2}-5x=-12$$

Ikki tarafini -3 ga bo‘ling.

$$\frac{-3x^{2}-5x}{-3}=-\frac{12}{-3}$$

-3 ga bo'lish -3 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{5}{-3}\right)x=-\frac{12}{-3}$$

-5 ni -3 ga bo'lish.

$$x^{2}+\frac{5}{3}x=-\frac{12}{-3}$$

-12 ni -3 ga bo'lish.

$$x^{2}+\frac{5}{3}x=4$$

\frac{5}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{5}{6} olish uchun. Keyin, \frac{5}{6} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{5}{3}x+\left(\frac{5}{6}\right)^{2}=4+\left(\frac{5}{6}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{5}{6} kvadratini chiqarish.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=4+\frac{25}{36}$$

4 ni \frac{25}{36} ga qo'shish.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=\frac{169}{36}$$

x^{2}+\frac{5}{3}x+\frac{25}{36} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{5}{6}\right)^{2}=\frac{169}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{5}{6}\right)^{2}}=\sqrt{\frac{169}{36}}$$

Qisqartirish.

$$x+\frac{5}{6}=\frac{13}{6} x+\frac{5}{6}=-\frac{13}{6}$$

Tenglamaning ikkala tarafidan \frac{5}{6} ni ayirish.

$$x=\frac{4}{3} x=-3$$