Whakaoti i te left(x+8right)^2=(x+2)y

$${ \left(x+8 \right) }^{ 2 } =(x+2)y$$

Whakaoti mō y

$$y=\frac{\left(x+8\right)^{2}}{x+2}$$

$$x\neq -2$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+8\right)^{2}.

$$x^{2}+16x+64=\left(x+2\right)y$$

Whakamahia te āhuatanga tohatoha hei whakarea te x+2 ki te y.

$$x^{2}+16x+64=xy+2y$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$xy+2y=x^{2}+16x+64$$

Pahekotia ngā kīanga tau katoa e whai ana i te y.

$$\left(x+2\right)y=x^{2}+16x+64$$

Whakawehea ngā taha e rua ki te x+2.

$$\frac{\left(x+2\right)y}{x+2}=\frac{\left(x+8\right)^{2}}{x+2}$$

Mā te whakawehe ki te x+2 ka wetekia te whakareanga ki te x+2.

$$y=\frac{\left(x+8\right)^{2}}{x+2}$$

Whakaoti mō x (complex solution)

$$x=\frac{\sqrt{y\left(y-24\right)}+y-16}{2}$$

$$x=\frac{-\sqrt{y\left(y-24\right)}+y-16}{2}$$

Whakaoti mō x

$$x=\frac{\sqrt{y\left(y-24\right)}+y-16}{2}$$

$$x=\frac{-\sqrt{y\left(y-24\right)}+y-16}{2}\text{, }y\geq 24\text{ or }y\leq 0$$