sqrt{5x+9}=2x+3 eching

$$\sqrt { 5 x + 9 } = 2 x + 3$$

x uchun yechish

$$x=0$$

Yechim qadamlari

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(\sqrt{5x+9}\right)^{2}=\left(2x+3\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{5x+9} ga hisoblang va 5x+9 ni qiymatni oling.

$$5x+9=\left(2x+3\right)^{2}$$

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

$$5x+9=4x^{2}+12x+9$$

Ikkala tarafdan 4x^{2} ni ayirish.

$$5x+9-4x^{2}=12x+9$$

Ikkala tarafdan 12x ni ayirish.

$$5x+9-4x^{2}-12x=9$$

-7x ni olish uchun 5x va -12x ni birlashtirish.

$$-7x+9-4x^{2}=9$$

Ikkala tarafdan 9 ni ayirish.

$$-7x+9-4x^{2}-9=0$$

0 olish uchun 9 dan 9 ni ayirish.

$$-7x-4x^{2}=0$$

x omili.

$$x\left(-7-4x\right)=0$$

Tenglamani yechish uchun x=0 va -7-4x=0 ni yeching.

$$x=0 x=-\frac{7}{4}$$

\sqrt{5x+9}=2x+3 tenglamasida x uchun 0 ni almashtiring.

$$\sqrt{5\times 0+9}=2\times 0+3$$

Qisqartirish. x=0 tenglamani qoniqtiradi.

$$3=3$$

\sqrt{5x+9}=2x+3 tenglamasida x uchun -\frac{7}{4} ni almashtiring.

$$\sqrt{5\left(-\frac{7}{4}\right)+9}=2\left(-\frac{7}{4}\right)+3$$

Qisqartirish. x=-\frac{7}{4} qiymati bu tenglamani qoniqtirmaydi, chunki oʻng va chap tarafdagi belgilar bir-biriga qarama-qarshi.

$$\frac{1}{2}=-\frac{1}{2}$$

\sqrt{5x+9}=2x+3 tenglamasi noyob yechimga ega.

$$x=0$$