2sqrt{x+5}=x+2? eching

$$2 \sqrt { x + 5 } = x + 2 ?$$

x uchun yechish

$$x=4$$

Yechim qadamlari

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(2\sqrt{x+5}\right)^{2}=\left(x+2\right)^{2}$$

\left(2\sqrt{x+5}\right)^{2} ni kengaytirish.

$$2^{2}\left(\sqrt{x+5}\right)^{2}=\left(x+2\right)^{2}$$

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

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

2 daraja ko‘rsatkichini \sqrt{x+5} ga hisoblang va x+5 ni qiymatni oling.

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

4 ga x+5 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$4x+20=\left(x+2\right)^{2}$$

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

$$4x+20=x^{2}+4x+4$$

Ikkala tarafdan x^{2} ni ayirish.

$$4x+20-x^{2}=4x+4$$

Ikkala tarafdan 4x ni ayirish.

$$4x+20-x^{2}-4x=4$$

0 ni olish uchun 4x va -4x ni birlashtirish.

$$20-x^{2}=4$$

Ikkala tarafdan 20 ni ayirish.

$$-x^{2}=4-20$$

-16 olish uchun 4 dan 20 ni ayirish.

$$-x^{2}=-16$$

Ikki tarafini -1 ga bo‘ling.

$$x^{2}=\frac{-16}{-1}$$

Ikkala surat va maxrajdan manfiy belgini olib tashlash bilan \frac{-16}{-1} kasrini 16 ga soddalashtirish mumkin.

$$x^{2}=16$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$x=4 x=-4$$

2\sqrt{x+5}=x+2 tenglamasida x uchun 4 ni almashtiring.

$$2\sqrt{4+5}=4+2$$

Qisqartirish. x=4 tenglamani qoniqtiradi.

$$6=6$$

2\sqrt{x+5}=x+2 tenglamasida x uchun -4 ni almashtiring.

$$2\sqrt{-4+5}=-4+2$$

Qisqartirish. x=-4 qiymati bu tenglamani qoniqtirmaydi, chunki oʻng va chap tarafdagi belgilar bir-biriga qarama-qarshi.

$$2=-2$$

2\sqrt{x+5}=x+2 tenglamasi noyob yechimga ega.

$$x=4$$