Whakaoti i te 1/4=(2k+2)^2-(1-x)/(8-k)^2

$$\frac { 1 } { 4 } = \frac { ( 2 k + 2 ) ^ { 2 } - ( 1 - x ) } { ( 8 - k ) ^ { 2 } }$$

Whakaoti mō x

$$x=-\frac{15k^{2}}{4}-12k+13$$

$$k\neq 8$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Me whakarea ngā taha e rua o te whārite ki te 4\left(k-8\right)^{2}, arā, te tauraro pātahi he tino iti rawa te kitea o 4,\left(8-k\right)^{2}.

$$\left(k-8\right)^{2}=4\left(\left(2k+2\right)^{2}-\left(1-x\right)\right)$$

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

$$k^{2}-16k+64=4\left(\left(2k+2\right)^{2}-\left(1-x\right)\right)$$

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

$$k^{2}-16k+64=4\left(4k^{2}+8k+4-\left(1-x\right)\right)$$

Hei kimi i te tauaro o 1-x, kimihia te tauaro o ia taurangi.

$$k^{2}-16k+64=4\left(4k^{2}+8k+4-1+x\right)$$

Tangohia te 1 i te 4, ka 3.

$$k^{2}-16k+64=4\left(4k^{2}+8k+3+x\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te 4k^{2}+8k+3+x.

$$k^{2}-16k+64=16k^{2}+32k+12+4x$$

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

$$16k^{2}+32k+12+4x=k^{2}-16k+64$$

Tangohia te 16k^{2} mai i ngā taha e rua.

$$32k+12+4x=k^{2}-16k+64-16k^{2}$$

Pahekotia te k^{2} me -16k^{2}, ka -15k^{2}.

$$32k+12+4x=-15k^{2}-16k+64$$

Tangohia te 32k mai i ngā taha e rua.

$$12+4x=-15k^{2}-16k+64-32k$$

Pahekotia te -16k me -32k, ka -48k.

$$12+4x=-15k^{2}-48k+64$$

Tangohia te 12 mai i ngā taha e rua.

$$4x=-15k^{2}-48k+64-12$$

Tangohia te 12 i te 64, ka 52.

$$4x=-15k^{2}-48k+52$$

He hanga arowhānui tō te whārite.

$$4x=52-48k-15k^{2}$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4x}{4}=\frac{52-48k-15k^{2}}{4}$$

Mā te whakawehe ki te 4 ka wetekia te whakareanga ki te 4.

$$x=\frac{52-48k-15k^{2}}{4}$$

Whakawehe -15k^{2}-48k+52 ki te 4.

$$x=-\frac{15k^{2}}{4}-12k+13$$

Whakaoti mō k (complex solution)

$$\left\{\begin{matrix}\\k=-\frac{2\sqrt{339-15x}}{15}-\frac{8}{5}\text{, }&\text{unconditionally}\\k=\frac{2\sqrt{339-15x}}{15}-\frac{8}{5}\text{, }&x\neq -323\end{matrix}\right.$$

Whakaoti mō k

$$\left\{\begin{matrix}k=\frac{2\sqrt{339-15x}}{15}-\frac{8}{5}\text{, }&x\neq -323\text{ and }x\leq \frac{113}{5}\\k=-\frac{2\sqrt{339-15x}}{15}-\frac{8}{5}\text{, }&x\leq \frac{113}{5}\end{matrix}\right.$$