1-(3x+2)^2=0 eching

$$1 - ( 3 x + 2 ) ^ { 2 } = 0$$

x uchun yechish

$$x=-\frac{1}{3}\approx -0,333333333$$

$$x=-1$$

Guruhlash orqali faktorlash qadamlari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(3x+2\right)^{2} kengaytirilishi uchun ishlating.

$$1-\left(9x^{2}+12x+4\right)=0$$

9x^{2}+12x+4 teskarisini topish uchun har birining teskarisini toping.

$$1-9x^{2}-12x-4=0$$

-3 olish uchun 1 dan 4 ni ayirish.

$$-3-9x^{2}-12x=0$$

Ikki tarafini 3 ga bo‘ling.

$$-1-3x^{2}-4x=0$$

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

$$-3x^{2}-4x-1=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon -3x^{2}+ax+bx-1 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-4 ab=-3\left(-1\right)=3$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. Faqat bundan juftlik tizim yechimidir.

$$a=-1 b=-3$$

-3x^{2}-4x-1 ni \left(-3x^{2}-x\right)+\left(-3x-1\right) sifatida qaytadan yozish.

$$\left(-3x^{2}-x\right)+\left(-3x-1\right)$$

Birinchi guruhda -x ni va ikkinchi guruhda -1 ni faktordan chiqaring.

$$-x\left(3x+1\right)-\left(3x+1\right)$$

Distributiv funktsiyasidan foydalangan holda 3x+1 umumiy terminini chiqaring.

$$\left(3x+1\right)\left(-x-1\right)$$

Tenglamani yechish uchun 3x+1=0 va -x-1=0 ni yeching.

$$x=-\frac{1}{3} x=-1$$

Kvadrat tenglamalaridan foydalanish qoidalari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(3x+2\right)^{2} kengaytirilishi uchun ishlating.

$$1-\left(9x^{2}+12x+4\right)=0$$

9x^{2}+12x+4 teskarisini topish uchun har birining teskarisini toping.

$$1-9x^{2}-12x-4=0$$

-3 olish uchun 1 dan 4 ni ayirish.

$$-3-9x^{2}-12x=0$$

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.

$$-9x^{2}-12x-3=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -9 ni a, -12 ni b va -3 ni c bilan almashtiring.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-9\right)\left(-3\right)}}{2\left(-9\right)}$$

-12 kvadratini chiqarish.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\left(-9\right)\left(-3\right)}}{2\left(-9\right)}$$

-4 ni -9 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144+36\left(-3\right)}}{2\left(-9\right)}$$

36 ni -3 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144-108}}{2\left(-9\right)}$$

144 ni -108 ga qo'shish.

$$x=\frac{-\left(-12\right)±\sqrt{36}}{2\left(-9\right)}$$

36 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-12\right)±6}{2\left(-9\right)}$$

-12 ning teskarisi 12 ga teng.

$$x=\frac{12±6}{2\left(-9\right)}$$

2 ni -9 marotabaga ko'paytirish.

$$x=\frac{12±6}{-18}$$

x=\frac{12±6}{-18} tenglamasini yeching, bunda ± musbat. 12 ni 6 ga qo'shish.

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

18 ni -18 ga bo'lish.

$$x=-1$$

x=\frac{12±6}{-18} tenglamasini yeching, bunda ± manfiy. 12 dan 6 ni ayirish.

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

\frac{6}{-18} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{1}{3}$$

Tenglama yechildi.

$$x=-1 x=-\frac{1}{3}$$

Kvadratni yakunlash bo'yicha qadamlar

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(3x+2\right)^{2} kengaytirilishi uchun ishlating.

$$1-\left(9x^{2}+12x+4\right)=0$$

9x^{2}+12x+4 teskarisini topish uchun har birining teskarisini toping.

$$1-9x^{2}-12x-4=0$$

-3 olish uchun 1 dan 4 ni ayirish.

$$-3-9x^{2}-12x=0$$

3 ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

$$-9x^{2}-12x=3$$

Ikki tarafini -9 ga bo‘ling.

$$\frac{-9x^{2}-12x}{-9}=\frac{3}{-9}$$

-9 ga bo'lish -9 ga ko'paytirishni bekor qiladi.

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

\frac{-12}{-9} ulushini 3 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{4}{3}x=\frac{3}{-9}$$

\frac{3}{-9} ulushini 3 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{4}{3}x=-\frac{1}{3}$$

\frac{4}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{2}{3} olish uchun. Keyin, \frac{2}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{4}{3}x+\left(\frac{2}{3}\right)^{2}=-\frac{1}{3}+\left(\frac{2}{3}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{2}{3} kvadratini chiqarish.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=-\frac{1}{3}+\frac{4}{9}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{1}{3} ni \frac{4}{9} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{1}{9}$$

x^{2}+\frac{4}{3}x+\frac{4}{9} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{2}{3}\right)^{2}=\frac{1}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{2}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Qisqartirish.

$$x+\frac{2}{3}=\frac{1}{3} x+\frac{2}{3}=-\frac{1}{3}$$

Tenglamaning ikkala tarafidan \frac{2}{3} ni ayirish.

$$x=-\frac{1}{3} x=-1$$