(2x+1)^2+(2x+3)^2=34 eching

$$( 2 x + 1 ) ^ { 2 } + ( 2 x + 3 ) ^ { 2 } = 34$$

x uchun yechish

$$x=-3$$

$$x=1$$

Guruhlash orqali faktorlash qadamlari

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

$$4x^{2}+4x+1+\left(2x+3\right)^{2}=34$$

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

$$4x^{2}+4x+1+4x^{2}+12x+9=34$$

8x^{2} ni olish uchun 4x^{2} va 4x^{2} ni birlashtirish.

$$8x^{2}+4x+1+12x+9=34$$

16x ni olish uchun 4x va 12x ni birlashtirish.

$$8x^{2}+16x+1+9=34$$

10 olish uchun 1 va 9'ni qo'shing.

$$8x^{2}+16x+10=34$$

Ikkala tarafdan 34 ni ayirish.

$$8x^{2}+16x+10-34=0$$

-24 olish uchun 10 dan 34 ni ayirish.

$$8x^{2}+16x-24=0$$

Ikki tarafini 8 ga bo‘ling.

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

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

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

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.

$$a=-1 b=3$$

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

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

Birinchi guruhda x ni va ikkinchi guruhda 3 ni faktordan chiqaring.

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

Distributiv funktsiyasidan foydalangan holda x-1 umumiy terminini chiqaring.

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

Tenglamani yechish uchun x-1=0 va x+3=0 ni yeching.

$$x=1 x=-3$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

$$4x^{2}+4x+1+\left(2x+3\right)^{2}=34$$

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

$$4x^{2}+4x+1+4x^{2}+12x+9=34$$

8x^{2} ni olish uchun 4x^{2} va 4x^{2} ni birlashtirish.

$$8x^{2}+4x+1+12x+9=34$$

16x ni olish uchun 4x va 12x ni birlashtirish.

$$8x^{2}+16x+1+9=34$$

10 olish uchun 1 va 9'ni qo'shing.

$$8x^{2}+16x+10=34$$

Ikkala tarafdan 34 ni ayirish.

$$8x^{2}+16x+10-34=0$$

-24 olish uchun 10 dan 34 ni ayirish.

$$8x^{2}+16x-24=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 8 ni a, 16 ni b va -24 ni c bilan almashtiring.

$$x=\frac{-16±\sqrt{16^{2}-4\times 8\left(-24\right)}}{2\times 8}$$

16 kvadratini chiqarish.

$$x=\frac{-16±\sqrt{256-4\times 8\left(-24\right)}}{2\times 8}$$

-4 ni 8 marotabaga ko'paytirish.

$$x=\frac{-16±\sqrt{256-32\left(-24\right)}}{2\times 8}$$

-32 ni -24 marotabaga ko'paytirish.

$$x=\frac{-16±\sqrt{256+768}}{2\times 8}$$

256 ni 768 ga qo'shish.

$$x=\frac{-16±\sqrt{1024}}{2\times 8}$$

1024 ning kvadrat ildizini chiqarish.

$$x=\frac{-16±32}{2\times 8}$$

2 ni 8 marotabaga ko'paytirish.

$$x=\frac{-16±32}{16}$$

x=\frac{-16±32}{16} tenglamasini yeching, bunda ± musbat. -16 ni 32 ga qo'shish.

$$x=\frac{16}{16}$$

16 ni 16 ga bo'lish.

$$x=1$$

x=\frac{-16±32}{16} tenglamasini yeching, bunda ± manfiy. -16 dan 32 ni ayirish.

$$x=-\frac{48}{16}$$

-48 ni 16 ga bo'lish.

$$x=-3$$

Tenglama yechildi.

$$x=1 x=-3$$

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.

$$4x^{2}+4x+1+\left(2x+3\right)^{2}=34$$

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

$$4x^{2}+4x+1+4x^{2}+12x+9=34$$

8x^{2} ni olish uchun 4x^{2} va 4x^{2} ni birlashtirish.

$$8x^{2}+4x+1+12x+9=34$$

16x ni olish uchun 4x va 12x ni birlashtirish.

$$8x^{2}+16x+1+9=34$$

10 olish uchun 1 va 9'ni qo'shing.

$$8x^{2}+16x+10=34$$

Ikkala tarafdan 10 ni ayirish.

$$8x^{2}+16x=34-10$$

24 olish uchun 34 dan 10 ni ayirish.

$$8x^{2}+16x=24$$

Ikki tarafini 8 ga bo‘ling.

$$\frac{8x^{2}+16x}{8}=\frac{24}{8}$$

8 ga bo'lish 8 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{16}{8}x=\frac{24}{8}$$

16 ni 8 ga bo'lish.

$$x^{2}+2x=\frac{24}{8}$$

24 ni 8 ga bo'lish.

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

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

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

1 kvadratini chiqarish.

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

3 ni 1 ga qo'shish.

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

x^{2}+2x+1 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x+1=2 x+1=-2$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x=1 x=-3$$