left(u+1right)^2=2u^2+5u+3 eching

$${ \left( u+1 \right) }^{ 2 } =2 { u }^{ 2 } +5u+3$$

u uchun yechish

$$u=-1$$

$$u=-2$$

Guruhlash orqali faktorlash qadamlari

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

$$u^{2}+2u+1=2u^{2}+5u+3$$

Ikkala tarafdan 2u^{2} ni ayirish.

$$u^{2}+2u+1-2u^{2}=5u+3$$

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

$$-u^{2}+2u+1=5u+3$$

Ikkala tarafdan 5u ni ayirish.

$$-u^{2}+2u+1-5u=3$$

-3u ni olish uchun 2u va -5u ni birlashtirish.

$$-u^{2}-3u+1=3$$

Ikkala tarafdan 3 ni ayirish.

$$-u^{2}-3u+1-3=0$$

-2 olish uchun 1 dan 3 ni ayirish.

$$-u^{2}-3u-2=0$$

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

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

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. Faqat bundan juftlik tizim yechimidir.

$$a=-1 b=-2$$

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

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

Birinchi guruhda u ni va ikkinchi guruhda 2 ni faktordan chiqaring.

$$u\left(-u-1\right)+2\left(-u-1\right)$$

Distributiv funktsiyasidan foydalangan holda -u-1 umumiy terminini chiqaring.

$$\left(-u-1\right)\left(u+2\right)$$

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

$$u=-1 u=-2$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

$$u^{2}+2u+1=2u^{2}+5u+3$$

Ikkala tarafdan 2u^{2} ni ayirish.

$$u^{2}+2u+1-2u^{2}=5u+3$$

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

$$-u^{2}+2u+1=5u+3$$

Ikkala tarafdan 5u ni ayirish.

$$-u^{2}+2u+1-5u=3$$

-3u ni olish uchun 2u va -5u ni birlashtirish.

$$-u^{2}-3u+1=3$$

Ikkala tarafdan 3 ni ayirish.

$$-u^{2}-3u+1-3=0$$

-2 olish uchun 1 dan 3 ni ayirish.

$$-u^{2}-3u-2=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -1 ni a, -3 ni b va -2 ni c bilan almashtiring.

$$u=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\left(-2\right)}}{2\left(-1\right)}$$

-3 kvadratini chiqarish.

$$u=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\left(-2\right)}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$u=\frac{-\left(-3\right)±\sqrt{9+4\left(-2\right)}}{2\left(-1\right)}$$

4 ni -2 marotabaga ko'paytirish.

$$u=\frac{-\left(-3\right)±\sqrt{9-8}}{2\left(-1\right)}$$

9 ni -8 ga qo'shish.

$$u=\frac{-\left(-3\right)±\sqrt{1}}{2\left(-1\right)}$$

1 ning kvadrat ildizini chiqarish.

$$u=\frac{-\left(-3\right)±1}{2\left(-1\right)}$$

-3 ning teskarisi 3 ga teng.

$$u=\frac{3±1}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$u=\frac{3±1}{-2}$$

u=\frac{3±1}{-2} tenglamasini yeching, bunda ± musbat. 3 ni 1 ga qo'shish.

$$u=\frac{4}{-2}$$

4 ni -2 ga bo'lish.

$$u=-2$$

u=\frac{3±1}{-2} tenglamasini yeching, bunda ± manfiy. 3 dan 1 ni ayirish.

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

2 ni -2 ga bo'lish.

$$u=-1$$

Tenglama yechildi.

$$u=-2 u=-1$$

Kvadratni yakunlash bo'yicha qadamlar

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

$$u^{2}+2u+1=2u^{2}+5u+3$$

Ikkala tarafdan 2u^{2} ni ayirish.

$$u^{2}+2u+1-2u^{2}=5u+3$$

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

$$-u^{2}+2u+1=5u+3$$

Ikkala tarafdan 5u ni ayirish.

$$-u^{2}+2u+1-5u=3$$

-3u ni olish uchun 2u va -5u ni birlashtirish.

$$-u^{2}-3u+1=3$$

Ikkala tarafdan 1 ni ayirish.

$$-u^{2}-3u=3-1$$

2 olish uchun 3 dan 1 ni ayirish.

$$-u^{2}-3u=2$$

Ikki tarafini -1 ga bo‘ling.

$$\frac{-u^{2}-3u}{-1}=\frac{2}{-1}$$

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

$$u^{2}+\left(-\frac{3}{-1}\right)u=\frac{2}{-1}$$

-3 ni -1 ga bo'lish.

$$u^{2}+3u=\frac{2}{-1}$$

2 ni -1 ga bo'lish.

$$u^{2}+3u=-2$$

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.

$$u^{2}+3u+\left(\frac{3}{2}\right)^{2}=-2+\left(\frac{3}{2}\right)^{2}$$

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

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

-2 ni \frac{9}{4} ga qo'shish.

$$u^{2}+3u+\frac{9}{4}=\frac{1}{4}$$

u^{2}+3u+\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.

$$\left(u+\frac{3}{2}\right)^{2}=\frac{1}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(u+\frac{3}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Qisqartirish.

$$u+\frac{3}{2}=\frac{1}{2} u+\frac{3}{2}=-\frac{1}{2}$$

Tenglamaning ikkala tarafidan \frac{3}{2} ni ayirish.

$$u=-1 u=-2$$