(3x+1)^2=4 eching

$$( 3 x + 1 ) ^ { 2 } = 4$$

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+1\right)^{2} kengaytirilishi uchun ishlating.

$$9x^{2}+6x+1=4$$

Ikkala tarafdan 4 ni ayirish.

$$9x^{2}+6x+1-4=0$$

-3 olish uchun 1 dan 4 ni ayirish.

$$9x^{2}+6x-3=0$$

Ikki tarafini 3 ga bo‘ling.

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

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.

$$a+b=2 ab=3\left(-1\right)=-3$$

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

$$a=-1 b=3$$

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

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

3x^{2}-x ichida x ni ajrating.

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

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+1\right)^{2} kengaytirilishi uchun ishlating.

$$9x^{2}+6x+1=4$$

Ikkala tarafdan 4 ni ayirish.

$$9x^{2}+6x+1-4=0$$

-3 olish uchun 1 dan 4 ni ayirish.

$$9x^{2}+6x-3=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 9 ni a, 6 ni b va -3 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}-4\times 9\left(-3\right)}}{2\times 9}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\times 9\left(-3\right)}}{2\times 9}$$

-4 ni 9 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-36\left(-3\right)}}{2\times 9}$$

-36 ni -3 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+108}}{2\times 9}$$

36 ni 108 ga qo'shish.

$$x=\frac{-6±\sqrt{144}}{2\times 9}$$

144 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±12}{2\times 9}$$

2 ni 9 marotabaga ko'paytirish.

$$x=\frac{-6±12}{18}$$

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

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

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

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

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

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

-18 ni 18 ga bo'lish.

$$x=-1$$

Tenglama yechildi.

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

Kvadratni yakunlash bo'yicha qadamlar

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(3x+1\right)^{2} kengaytirilishi uchun ishlating.

$$9x^{2}+6x+1=4$$

Ikkala tarafdan 1 ni ayirish.

$$9x^{2}+6x=4-1$$

3 olish uchun 4 dan 1 ni ayirish.

$$9x^{2}+6x=3$$

Ikki tarafini 9 ga bo‘ling.

$$\frac{9x^{2}+6x}{9}=\frac{3}{9}$$

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

$$x^{2}+\frac{6}{9}x=\frac{3}{9}$$

\frac{6}{9} ulushini 3 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

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

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

\frac{2}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{3} olish uchun. Keyin, \frac{1}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{3} kvadratini chiqarish.

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

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

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

x^{2}+\frac{2}{3}x+\frac{1}{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.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{4}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Qisqartirish.

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

Tenglamaning ikkala tarafidan \frac{1}{3} ni ayirish.

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