Réitigh x^2+2sqrt{2}kx+18=0

$${ x }^{ 2 } +2 \sqrt{ 2 } kx+18=0$$

Réitigh do k.

$$k=-\frac{\sqrt{2}\left(x^{2}+18\right)}{4x}$$

$$x\neq 0$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Bain x^{2} ón dá thaobh. Is ionann rud ar bith a dhealaítear ó nialas agus a shéanadh.

$$2\sqrt{2}kx+18=-x^{2}$$

Bain 18 ón dá thaobh.

$$2\sqrt{2}kx=-x^{2}-18$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$2\sqrt{2}xk=-x^{2}-18$$

Roinn an dá thaobh faoi 2\sqrt{2}x.

$$\frac{2\sqrt{2}xk}{2\sqrt{2}x}=\frac{-x^{2}-18}{2\sqrt{2}x}$$

Má roinntear é faoi 2\sqrt{2}x cuirtear an iolrúchán faoi 2\sqrt{2}x ar ceal.

$$k=\frac{-x^{2}-18}{2\sqrt{2}x}$$

Roinn -x^{2}-18 faoi 2\sqrt{2}x.

$$k=-\frac{\sqrt{2}\left(x^{2}+18\right)}{4x}$$

Réitigh do x. (complex solution)

$$x=\sqrt{2}\left(\sqrt{k^{2}-9}-k\right)$$

$$x=\sqrt{2}\left(-\sqrt{k^{2}-9}-k\right)$$

Réitigh do x.

$$x=\sqrt{2}\left(\sqrt{k^{2}-9}-k\right)$$

$$x=\sqrt{2}\left(-\sqrt{k^{2}-9}-k\right)\text{, }|k|\geq 3$$