(2x-3)^2=16 eching

$$( 2 x - 3 ) ^ { 2 } = 16$$

x uchun yechish

$$x = \frac{7}{2} = 3\frac{1}{2} = 3,5$$

$$x=-\frac{1}{2}=-0,5$$

Guruhlash orqali faktorlash qadamlari

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

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

Ikkala tarafdan 16 ni ayirish.

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

-7 olish uchun 9 dan 16 ni ayirish.

$$4x^{2}-12x-7=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 4x^{2}+ax+bx-7 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-12 ab=4\left(-7\right)=-28$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -28-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-28 2,-14 4,-7$$

Har bir juftlik yigʻindisini hisoblang.

$$1-28=-27 2-14=-12 4-7=-3$$

Yechim – -12 yigʻindisini beruvchi juftlik.

$$a=-14 b=2$$

4x^{2}-12x-7 ni \left(4x^{2}-14x\right)+\left(2x-7\right) sifatida qaytadan yozish.

$$\left(4x^{2}-14x\right)+\left(2x-7\right)$$

4x^{2}-14x ichida 2x ni ajrating.

$$2x\left(2x-7\right)+2x-7$$

Distributiv funktsiyasidan foydalangan holda 2x-7 umumiy terminini chiqaring.

$$\left(2x-7\right)\left(2x+1\right)$$

Tenglamani yechish uchun 2x-7=0 va 2x+1=0 ni yeching.

$$x=\frac{7}{2} x=-\frac{1}{2}$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

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

Ikkala tarafdan 16 ni ayirish.

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

-7 olish uchun 9 dan 16 ni ayirish.

$$4x^{2}-12x-7=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 4 ni a, -12 ni b va -7 ni c bilan almashtiring.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\left(-7\right)}}{2\times 4}$$

-12 kvadratini chiqarish.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 4\left(-7\right)}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144-16\left(-7\right)}}{2\times 4}$$

-16 ni -7 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144+112}}{2\times 4}$$

144 ni 112 ga qo'shish.

$$x=\frac{-\left(-12\right)±\sqrt{256}}{2\times 4}$$

256 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-12\right)±16}{2\times 4}$$

-12 ning teskarisi 12 ga teng.

$$x=\frac{12±16}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{12±16}{8}$$

x=\frac{12±16}{8} tenglamasini yeching, bunda ± musbat. 12 ni 16 ga qo'shish.

$$x=\frac{28}{8}$$

\frac{28}{8} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{7}{2}$$

x=\frac{12±16}{8} tenglamasini yeching, bunda ± manfiy. 12 dan 16 ni ayirish.

$$x=-\frac{4}{8}$$

\frac{-4}{8} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

Tenglama yechildi.

$$x=\frac{7}{2} x=-\frac{1}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

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

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

Ikkala tarafdan 9 ni ayirish.

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

7 olish uchun 16 dan 9 ni ayirish.

$$4x^{2}-12x=7$$

Ikki tarafini 4 ga bo‘ling.

$$\frac{4x^{2}-12x}{4}=\frac{7}{4}$$

4 ga bo'lish 4 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{12}{4}\right)x=\frac{7}{4}$$

-12 ni 4 ga bo'lish.

$$x^{2}-3x=\frac{7}{4}$$

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

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=\frac{7}{4}+\left(-\frac{3}{2}\right)^{2}$$

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

$$x^{2}-3x+\frac{9}{4}=\frac{7+9}{4}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{7}{4} ni \frac{9}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

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

x^{2}-3x+\frac{9}{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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

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

\frac{3}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{7}{2} x=-\frac{1}{2}$$