y=sqrt{5x+10} eching

$$y= \sqrt{ 5x+10 }$$

x uchun yechish

$$x=\frac{y^{2}-10}{5}$$

$$y\geq 0$$

Ildizli tenglamalarni yechish qadamlari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$\sqrt{5x+10}=y$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$5x+10=y^{2}$$

Tenglamaning ikkala tarafidan 10 ni ayirish.

$$5x+10-10=y^{2}-10$$

O‘zidan 10 ayirilsa 0 qoladi.

$$5x=y^{2}-10$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x}{5}=\frac{y^{2}-10}{5}$$

5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.

$$x=\frac{y^{2}-10}{5}$$

y^{2}-10 ni 5 ga bo'lish.

$$x=\frac{y^{2}}{5}-2$$

x uchun yechish (complex solution)

$$x=\frac{y^{2}-10}{5}$$

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

y uchun yechish (complex solution)

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

y uchun yechish

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

$$x\geq -2$$