(v+4)^2=2v^2+2v+9 eching

$$( v + 4 ) ^ { 2 } = 2 v ^ { 2 } + 2 v + 9$$

v uchun yechish

$$v=-1$$

$$v=7$$

Guruhlash orqali faktorlash qadamlari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(v+4\right)^{2} kengaytirilishi uchun ishlating.

$$v^{2}+8v+16=2v^{2}+2v+9$$

Ikkala tarafdan 2v^{2} ni ayirish.

$$v^{2}+8v+16-2v^{2}=2v+9$$

-v^{2} ni olish uchun v^{2} va -2v^{2} ni birlashtirish.

$$-v^{2}+8v+16=2v+9$$

Ikkala tarafdan 2v ni ayirish.

$$-v^{2}+8v+16-2v=9$$

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

$$-v^{2}+6v+16=9$$

Ikkala tarafdan 9 ni ayirish.

$$-v^{2}+6v+16-9=0$$

7 olish uchun 16 dan 9 ni ayirish.

$$-v^{2}+6v+7=0$$

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

$$a+b=6 ab=-7=-7$$

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.

$$a=7 b=-1$$

-v^{2}+6v+7 ni \left(-v^{2}+7v\right)+\left(-v+7\right) sifatida qaytadan yozish.

$$\left(-v^{2}+7v\right)+\left(-v+7\right)$$

Birinchi guruhda -v ni va ikkinchi guruhda -1 ni faktordan chiqaring.

$$-v\left(v-7\right)-\left(v-7\right)$$

Distributiv funktsiyasidan foydalangan holda v-7 umumiy terminini chiqaring.

$$\left(v-7\right)\left(-v-1\right)$$

Tenglamani yechish uchun v-7=0 va -v-1=0 ni yeching.

$$v=7 v=-1$$

Kvadrat tenglamalaridan foydalanish qoidalari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(v+4\right)^{2} kengaytirilishi uchun ishlating.

$$v^{2}+8v+16=2v^{2}+2v+9$$

Ikkala tarafdan 2v^{2} ni ayirish.

$$v^{2}+8v+16-2v^{2}=2v+9$$

-v^{2} ni olish uchun v^{2} va -2v^{2} ni birlashtirish.

$$-v^{2}+8v+16=2v+9$$

Ikkala tarafdan 2v ni ayirish.

$$-v^{2}+8v+16-2v=9$$

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

$$-v^{2}+6v+16=9$$

Ikkala tarafdan 9 ni ayirish.

$$-v^{2}+6v+16-9=0$$

7 olish uchun 16 dan 9 ni ayirish.

$$-v^{2}+6v+7=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -1 ni a, 6 ni b va 7 ni c bilan almashtiring.

$$v=\frac{-6±\sqrt{6^{2}-4\left(-1\right)\times 7}}{2\left(-1\right)}$$

6 kvadratini chiqarish.

$$v=\frac{-6±\sqrt{36-4\left(-1\right)\times 7}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$v=\frac{-6±\sqrt{36+4\times 7}}{2\left(-1\right)}$$

4 ni 7 marotabaga ko'paytirish.

$$v=\frac{-6±\sqrt{36+28}}{2\left(-1\right)}$$

36 ni 28 ga qo'shish.

$$v=\frac{-6±\sqrt{64}}{2\left(-1\right)}$$

64 ning kvadrat ildizini chiqarish.

$$v=\frac{-6±8}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$v=\frac{-6±8}{-2}$$

v=\frac{-6±8}{-2} tenglamasini yeching, bunda ± musbat. -6 ni 8 ga qo'shish.

$$v=\frac{2}{-2}$$

2 ni -2 ga bo'lish.

$$v=-1$$

v=\frac{-6±8}{-2} tenglamasini yeching, bunda ± manfiy. -6 dan 8 ni ayirish.

$$v=-\frac{14}{-2}$$

-14 ni -2 ga bo'lish.

$$v=7$$

Tenglama yechildi.

$$v=-1 v=7$$

Kvadratni yakunlash bo'yicha qadamlar

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(v+4\right)^{2} kengaytirilishi uchun ishlating.

$$v^{2}+8v+16=2v^{2}+2v+9$$

Ikkala tarafdan 2v^{2} ni ayirish.

$$v^{2}+8v+16-2v^{2}=2v+9$$

-v^{2} ni olish uchun v^{2} va -2v^{2} ni birlashtirish.

$$-v^{2}+8v+16=2v+9$$

Ikkala tarafdan 2v ni ayirish.

$$-v^{2}+8v+16-2v=9$$

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

$$-v^{2}+6v+16=9$$

Ikkala tarafdan 16 ni ayirish.

$$-v^{2}+6v=9-16$$

-7 olish uchun 9 dan 16 ni ayirish.

$$-v^{2}+6v=-7$$

Ikki tarafini -1 ga bo‘ling.

$$\frac{-v^{2}+6v}{-1}=-\frac{7}{-1}$$

-1 ga bo'lish -1 ga ko'paytirishni bekor qiladi.

$$v^{2}+\frac{6}{-1}v=-\frac{7}{-1}$$

6 ni -1 ga bo'lish.

$$v^{2}-6v=-\frac{7}{-1}$$

-7 ni -1 ga bo'lish.

$$v^{2}-6v=7$$

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

$$v^{2}-6v+\left(-3\right)^{2}=7+\left(-3\right)^{2}$$

-3 kvadratini chiqarish.

$$v^{2}-6v+9=7+9$$

7 ni 9 ga qo'shish.

$$v^{2}-6v+9=16$$

v^{2}-6v+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.

$$\left(v-3\right)^{2}=16$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(v-3\right)^{2}}=\sqrt{16}$$

Qisqartirish.

$$v-3=4 v-3=-4$$

3 ni tenglamaning ikkala tarafiga qo'shish.

$$v=7 v=-1$$