y=sqrt{1/4x+2} eching

$$y = \sqrt{ \frac{ 1 }{ 4 } x+2 }$$

x uchun yechish

$$x=4\left(y^{2}-2\right)$$

$$y\geq 0$$

Ildizli tenglamalarni yechish qadamlari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$\sqrt{\frac{1}{4}x+2}=y$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\frac{1}{4}x+2=y^{2}$$

Tenglamaning ikkala tarafidan 2 ni ayirish.

$$\frac{1}{4}x+2-2=y^{2}-2$$

O‘zidan 2 ayirilsa 0 qoladi.

$$\frac{1}{4}x=y^{2}-2$$

Ikkala tarafini 4 ga ko‘paytiring.

$$\frac{\frac{1}{4}x}{\frac{1}{4}}=\frac{y^{2}-2}{\frac{1}{4}}$$

\frac{1}{4} ga bo'lish \frac{1}{4} ga ko'paytirishni bekor qiladi.

$$x=\frac{y^{2}-2}{\frac{1}{4}}$$

y^{2}-2 ni \frac{1}{4} ga bo'lish y^{2}-2 ga k'paytirish \frac{1}{4} ga qaytarish.

$$x=4y^{2}-8$$

x uchun yechish (complex solution)

$$x=4\left(y^{2}-2\right)$$

$$arg(y)<\pi \text{ or }y=0$$

y uchun yechish (complex solution)

$$y=\frac{\sqrt{x+8}}{2}$$

y uchun yechish

$$y=\frac{\sqrt{x+8}}{2}$$

$$x\geq -8$$