(2x+1)^2=(x-5)^2 eching

$$( 2 x + 1 ) ^ { 2 } = ( x - 5 ) ^ { 2 }$$

x uchun yechish

$$x=-6$$

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

Guruhlash orqali faktorlash qadamlari

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

$$4x^{2}+4x+1=\left(x-5\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-5\right)^{2} kengaytirilishi uchun ishlating.

$$4x^{2}+4x+1=x^{2}-10x+25$$

Ikkala tarafdan x^{2} ni ayirish.

$$4x^{2}+4x+1-x^{2}=-10x+25$$

3x^{2} ni olish uchun 4x^{2} va -x^{2} ni birlashtirish.

$$3x^{2}+4x+1=-10x+25$$

10x ni ikki tarafga qo’shing.

$$3x^{2}+4x+1+10x=25$$

14x ni olish uchun 4x va 10x ni birlashtirish.

$$3x^{2}+14x+1=25$$

Ikkala tarafdan 25 ni ayirish.

$$3x^{2}+14x+1-25=0$$

-24 olish uchun 1 dan 25 ni ayirish.

$$3x^{2}+14x-24=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 3x^{2}+ax+bx-24 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=14 ab=3\left(-24\right)=-72$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -72-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

Yechim – 14 yigʻindisini beruvchi juftlik.

$$a=-4 b=18$$

3x^{2}+14x-24 ni \left(3x^{2}-4x\right)+\left(18x-24\right) sifatida qaytadan yozish.

$$\left(3x^{2}-4x\right)+\left(18x-24\right)$$

Birinchi guruhda x ni va ikkinchi guruhda 6 ni faktordan chiqaring.

$$x\left(3x-4\right)+6\left(3x-4\right)$$

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

$$\left(3x-4\right)\left(x+6\right)$$

Tenglamani yechish uchun 3x-4=0 va x+6=0 ni yeching.

$$x=\frac{4}{3} x=-6$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

$$4x^{2}+4x+1=\left(x-5\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-5\right)^{2} kengaytirilishi uchun ishlating.

$$4x^{2}+4x+1=x^{2}-10x+25$$

Ikkala tarafdan x^{2} ni ayirish.

$$4x^{2}+4x+1-x^{2}=-10x+25$$

3x^{2} ni olish uchun 4x^{2} va -x^{2} ni birlashtirish.

$$3x^{2}+4x+1=-10x+25$$

10x ni ikki tarafga qo’shing.

$$3x^{2}+4x+1+10x=25$$

14x ni olish uchun 4x va 10x ni birlashtirish.

$$3x^{2}+14x+1=25$$

Ikkala tarafdan 25 ni ayirish.

$$3x^{2}+14x+1-25=0$$

-24 olish uchun 1 dan 25 ni ayirish.

$$3x^{2}+14x-24=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, 14 ni b va -24 ni c bilan almashtiring.

$$x=\frac{-14±\sqrt{14^{2}-4\times 3\left(-24\right)}}{2\times 3}$$

14 kvadratini chiqarish.

$$x=\frac{-14±\sqrt{196-4\times 3\left(-24\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-14±\sqrt{196-12\left(-24\right)}}{2\times 3}$$

-12 ni -24 marotabaga ko'paytirish.

$$x=\frac{-14±\sqrt{196+288}}{2\times 3}$$

196 ni 288 ga qo'shish.

$$x=\frac{-14±\sqrt{484}}{2\times 3}$$

484 ning kvadrat ildizini chiqarish.

$$x=\frac{-14±22}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-14±22}{6}$$

x=\frac{-14±22}{6} tenglamasini yeching, bunda ± musbat. -14 ni 22 ga qo'shish.

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

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

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

x=\frac{-14±22}{6} tenglamasini yeching, bunda ± manfiy. -14 dan 22 ni ayirish.

$$x=-\frac{36}{6}$$

-36 ni 6 ga bo'lish.

$$x=-6$$

Tenglama yechildi.

$$x=\frac{4}{3} x=-6$$

Kvadratni yakunlash bo'yicha qadamlar

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

$$4x^{2}+4x+1=\left(x-5\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-5\right)^{2} kengaytirilishi uchun ishlating.

$$4x^{2}+4x+1=x^{2}-10x+25$$

Ikkala tarafdan x^{2} ni ayirish.

$$4x^{2}+4x+1-x^{2}=-10x+25$$

3x^{2} ni olish uchun 4x^{2} va -x^{2} ni birlashtirish.

$$3x^{2}+4x+1=-10x+25$$

10x ni ikki tarafga qo’shing.

$$3x^{2}+4x+1+10x=25$$

14x ni olish uchun 4x va 10x ni birlashtirish.

$$3x^{2}+14x+1=25$$

Ikkala tarafdan 1 ni ayirish.

$$3x^{2}+14x=25-1$$

24 olish uchun 25 dan 1 ni ayirish.

$$3x^{2}+14x=24$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}+14x}{3}=\frac{24}{3}$$

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

$$x^{2}+\frac{14}{3}x=\frac{24}{3}$$

24 ni 3 ga bo'lish.

$$x^{2}+\frac{14}{3}x=8$$

\frac{14}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{7}{3} olish uchun. Keyin, \frac{7}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{14}{3}x+\left(\frac{7}{3}\right)^{2}=8+\left(\frac{7}{3}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{7}{3} kvadratini chiqarish.

$$x^{2}+\frac{14}{3}x+\frac{49}{9}=8+\frac{49}{9}$$

8 ni \frac{49}{9} ga qo'shish.

$$x^{2}+\frac{14}{3}x+\frac{49}{9}=\frac{121}{9}$$

x^{2}+\frac{14}{3}x+\frac{49}{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.

$$\left(x+\frac{7}{3}\right)^{2}=\frac{121}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{7}{3}\right)^{2}}=\sqrt{\frac{121}{9}}$$

Qisqartirish.

$$x+\frac{7}{3}=\frac{11}{3} x+\frac{7}{3}=-\frac{11}{3}$$

Tenglamaning ikkala tarafidan \frac{7}{3} ni ayirish.

$$x=\frac{4}{3} x=-6$$