(1)(3x+2)^2=16 eching

$$( 1 ) ( 3 x + 2 ) ^ { 2 } = 16$$

x uchun yechish

$$x=\frac{2}{3}\approx 0,666666667$$

$$x=-2$$

Guruhlash orqali faktorlash qadamlari

Ikki tarafini 1 ga bo‘ling.

$$\left(3x+2\right)^{2}=16$$

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

$$9x^{2}+12x+4=16$$

Ikkala tarafdan 16 ni ayirish.

$$9x^{2}+12x+4-16=0$$

-12 olish uchun 4 dan 16 ni ayirish.

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

Ikki tarafini 3 ga bo‘ling.

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

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 3x^{2}+ax+bx-4 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=4 ab=3\left(-4\right)=-12$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -12-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,12 -2,6 -3,4$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+12=11 -2+6=4 -3+4=1$$

Yechim – 4 yigʻindisini beruvchi juftlik.

$$a=-2 b=6$$

3x^{2}+4x-4 ni \left(3x^{2}-2x\right)+\left(6x-4\right) sifatida qaytadan yozish.

$$\left(3x^{2}-2x\right)+\left(6x-4\right)$$

Birinchi guruhda x ni va ikkinchi guruhda 2 ni faktordan chiqaring.

$$x\left(3x-2\right)+2\left(3x-2\right)$$

Distributiv funktsiyasidan foydalangan holda 3x-2 umumiy terminini chiqaring.

$$\left(3x-2\right)\left(x+2\right)$$

Tenglamani yechish uchun 3x-2=0 va x+2=0 ni yeching.

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

Kvadrat tenglamalaridan foydalanish qoidalari

Ikki tarafini 1 ga bo‘ling.

$$\left(3x+2\right)^{2}=16$$

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

$$9x^{2}+12x+4=16$$

Ikkala tarafdan 16 ni ayirish.

$$9x^{2}+12x+4-16=0$$

-12 olish uchun 4 dan 16 ni ayirish.

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

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 -12 ni c bilan almashtiring.

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

12 kvadratini chiqarish.

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

-4 ni 9 marotabaga ko'paytirish.

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

-36 ni -12 marotabaga ko'paytirish.

$$x=\frac{-12±\sqrt{144+432}}{2\times 9}$$

144 ni 432 ga qo'shish.

$$x=\frac{-12±\sqrt{576}}{2\times 9}$$

576 ning kvadrat ildizini chiqarish.

$$x=\frac{-12±24}{2\times 9}$$

2 ni 9 marotabaga ko'paytirish.

$$x=\frac{-12±24}{18}$$

x=\frac{-12±24}{18} tenglamasini yeching, bunda ± musbat. -12 ni 24 ga qo'shish.

$$x=\frac{12}{18}$$

\frac{12}{18} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{2}{3}$$

x=\frac{-12±24}{18} tenglamasini yeching, bunda ± manfiy. -12 dan 24 ni ayirish.

$$x=-\frac{36}{18}$$

-36 ni 18 ga bo'lish.

$$x=-2$$

Tenglama yechildi.

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

Kvadratni yakunlash bo'yicha qadamlar

Ikki tarafini 1 ga bo‘ling.

$$\left(3x+2\right)^{2}=16$$

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

$$9x^{2}+12x+4=16$$

Ikkala tarafdan 4 ni ayirish.

$$9x^{2}+12x=16-4$$

12 olish uchun 16 dan 4 ni ayirish.

$$9x^{2}+12x=12$$

Ikki tarafini 9 ga bo‘ling.

$$\frac{9x^{2}+12x}{9}=\frac{12}{9}$$

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

$$x^{2}+\frac{12}{9}x=\frac{12}{9}$$

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

$$x^{2}+\frac{4}{3}x=\frac{12}{9}$$

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

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

\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.

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

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

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

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

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{16}{9}$$

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.

$$\left(x+\frac{2}{3}\right)^{2}=\frac{16}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{2}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Qisqartirish.

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

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

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