Réitigh 1/4=(2k+2)^2-(1-x)/(8-k)^2

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

Réitigh do x.

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

$$k\neq 8$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Iolraigh an dá thaobh den chothromóid faoi 4\left(k-8\right)^{2}, an comhiolraí is lú de 4,\left(8-k\right)^{2}.

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

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(k-8\right)^{2} a leathnú.

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

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{2}=a^{2}+2ab+b^{2} chun \left(2k+2\right)^{2} a leathnú.

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

Chun an mhalairt ar 1-x a aimsiú, aimsigh an mhalairt ar gach téarma.

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

Dealaigh 1 ó 4 chun 3 a fháil.

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

Úsáid an t-airí dáileach chun 4 a mhéadú faoi 4k^{2}+8k+3+x.

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

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

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

Bain 16k^{2} ón dá thaobh.

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

Comhcheangail k^{2} agus -16k^{2} chun -15k^{2} a fháil.

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

Bain 32k ón dá thaobh.

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

Comhcheangail -16k agus -32k chun -48k a fháil.

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

Bain 12 ón dá thaobh.

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

Dealaigh 12 ó 64 chun 52 a fháil.

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

Tá an chothromóid i bhfoirm chaighdeánach.

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

Roinn an dá thaobh faoi 4.

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

Má roinntear é faoi 4 cuirtear an iolrúchán faoi 4 ar ceal.

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

Roinn -15k^{2}-48k+52 faoi 4.

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

Réitigh do k. (complex solution)

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

Réitigh do k.

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