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

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

x uchun yechish

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

$$8-2y\geq 0$$

Ildizli tenglamalarni yechish qadamlari

\frac{-1}{2} kasri manfiy belgini olib tashlash bilan -\frac{1}{2} sifatida qayta yozilishi mumkin.

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

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

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

Ikkala tarafdan 4 ni ayirish.

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

Ikkala tarafini -2 ga ko‘paytiring.

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

-\frac{1}{2} ga bo'lish -\frac{1}{2} ga ko'paytirishni bekor qiladi.

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

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

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

Tenglamaning ikkala taraf kvadratini chiqarish.

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

Tenglamaning ikkala tarafidan 2 ni ayirish.

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

O‘zidan 2 ayirilsa 0 qoladi.

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

x uchun yechish (complex solution)

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

$$y=4\text{ or }arg(8-2y)<\pi$$

y uchun yechish (complex solution)

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

y uchun yechish

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

$$x\geq -2$$