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

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

x uchun yechish

$$x = -\frac{11}{5} = -2\frac{1}{5} = -2,2$$

$$x=1$$

Guruhlash orqali faktorlash qadamlari

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

$$x^{2}-2x+1+\left(2x+2\right)^{2}=16$$

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

$$x^{2}-2x+1+4x^{2}+8x+4=16$$

5x^{2} ni olish uchun x^{2} va 4x^{2} ni birlashtirish.

$$5x^{2}-2x+1+8x+4=16$$

6x ni olish uchun -2x va 8x ni birlashtirish.

$$5x^{2}+6x+1+4=16$$

5 olish uchun 1 va 4'ni qo'shing.

$$5x^{2}+6x+5=16$$

Ikkala tarafdan 16 ni ayirish.

$$5x^{2}+6x+5-16=0$$

-11 olish uchun 5 dan 16 ni ayirish.

$$5x^{2}+6x-11=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 5x^{2}+ax+bx-11 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=6 ab=5\left(-11\right)=-55$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -55-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,55 -5,11$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+55=54 -5+11=6$$

Yechim – 6 yigʻindisini beruvchi juftlik.

$$a=-5 b=11$$

5x^{2}+6x-11 ni \left(5x^{2}-5x\right)+\left(11x-11\right) sifatida qaytadan yozish.

$$\left(5x^{2}-5x\right)+\left(11x-11\right)$$

Birinchi guruhda 5x ni va ikkinchi guruhda 11 ni faktordan chiqaring.

$$5x\left(x-1\right)+11\left(x-1\right)$$

Distributiv funktsiyasidan foydalangan holda x-1 umumiy terminini chiqaring.

$$\left(x-1\right)\left(5x+11\right)$$

Tenglamani yechish uchun x-1=0 va 5x+11=0 ni yeching.

$$x=1 x=-\frac{11}{5}$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

$$x^{2}-2x+1+\left(2x+2\right)^{2}=16$$

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

$$x^{2}-2x+1+4x^{2}+8x+4=16$$

5x^{2} ni olish uchun x^{2} va 4x^{2} ni birlashtirish.

$$5x^{2}-2x+1+8x+4=16$$

6x ni olish uchun -2x va 8x ni birlashtirish.

$$5x^{2}+6x+1+4=16$$

5 olish uchun 1 va 4'ni qo'shing.

$$5x^{2}+6x+5=16$$

Ikkala tarafdan 16 ni ayirish.

$$5x^{2}+6x+5-16=0$$

-11 olish uchun 5 dan 16 ni ayirish.

$$5x^{2}+6x-11=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 6 ni b va -11 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}-4\times 5\left(-11\right)}}{2\times 5}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\times 5\left(-11\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-20\left(-11\right)}}{2\times 5}$$

-20 ni -11 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+220}}{2\times 5}$$

36 ni 220 ga qo'shish.

$$x=\frac{-6±\sqrt{256}}{2\times 5}$$

256 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±16}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-6±16}{10}$$

x=\frac{-6±16}{10} tenglamasini yeching, bunda ± musbat. -6 ni 16 ga qo'shish.

$$x=\frac{10}{10}$$

10 ni 10 ga bo'lish.

$$x=1$$

x=\frac{-6±16}{10} tenglamasini yeching, bunda ± manfiy. -6 dan 16 ni ayirish.

$$x=-\frac{22}{10}$$

\frac{-22}{10} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{11}{5}$$

Tenglama yechildi.

$$x=1 x=-\frac{11}{5}$$

Kvadratni yakunlash bo'yicha qadamlar

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

$$x^{2}-2x+1+\left(2x+2\right)^{2}=16$$

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

$$x^{2}-2x+1+4x^{2}+8x+4=16$$

5x^{2} ni olish uchun x^{2} va 4x^{2} ni birlashtirish.

$$5x^{2}-2x+1+8x+4=16$$

6x ni olish uchun -2x va 8x ni birlashtirish.

$$5x^{2}+6x+1+4=16$$

5 olish uchun 1 va 4'ni qo'shing.

$$5x^{2}+6x+5=16$$

Ikkala tarafdan 5 ni ayirish.

$$5x^{2}+6x=16-5$$

11 olish uchun 16 dan 5 ni ayirish.

$$5x^{2}+6x=11$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}+6x}{5}=\frac{11}{5}$$

5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{6}{5}x=\frac{11}{5}$$

\frac{6}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{3}{5} olish uchun. Keyin, \frac{3}{5} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{6}{5}x+\left(\frac{3}{5}\right)^{2}=\frac{11}{5}+\left(\frac{3}{5}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{5} kvadratini chiqarish.

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=\frac{11}{5}+\frac{9}{25}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{11}{5} ni \frac{9}{25} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=\frac{64}{25}$$

x^{2}+\frac{6}{5}x+\frac{9}{25} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{3}{5}\right)^{2}=\frac{64}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{5}\right)^{2}}=\sqrt{\frac{64}{25}}$$

Qisqartirish.

$$x+\frac{3}{5}=\frac{8}{5} x+\frac{3}{5}=-\frac{8}{5}$$

Tenglamaning ikkala tarafidan \frac{3}{5} ni ayirish.

$$x=1 x=-\frac{11}{5}$$