Réitigh 16x^2+(2k+1)x+k-1=0

$$16 x ^ { 2 } + ( 2 k + 1 ) x + k - 1 = 0$$

Réitigh do k.

$$k=-\frac{16x^{2}+x-1}{2x+1}$$

$$x\neq -\frac{1}{2}$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Úsáid an t-airí dáileach chun 2k+1 a mhéadú faoi x.

$$16x^{2}+2kx+x+k-1=0$$

Bain 16x^{2} ón dá thaobh. Is ionann rud ar bith a dhealaítear ó nialas agus a shéanadh.

$$2kx+x+k-1=-16x^{2}$$

Bain x ón dá thaobh.

$$2kx+k-1=-16x^{2}-x$$

Cuir 1 leis an dá thaobh.

$$2kx+k=-16x^{2}-x+1$$

Comhcheangail na téarmaí ar fad ina bhfuil k.

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

Tá an chothromóid i bhfoirm chaighdeánach.

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

Roinn an dá thaobh faoi 2x+1.

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

Má roinntear é faoi 2x+1 cuirtear an iolrúchán faoi 2x+1 ar ceal.

$$k=\frac{1-x-16x^{2}}{2x+1}$$

Réitigh do x. (complex solution)

$$x=\frac{\sqrt{4k^{2}-60k+65}}{32}-\frac{k}{16}-\frac{1}{32}$$

$$x=-\frac{\sqrt{4k^{2}-60k+65}}{32}-\frac{k}{16}-\frac{1}{32}$$

Réitigh do x.

$$x=\frac{\sqrt{4k^{2}-60k+65}}{32}-\frac{k}{16}-\frac{1}{32}$$

$$x=-\frac{\sqrt{4k^{2}-60k+65}}{32}-\frac{k}{16}-\frac{1}{32}\text{, }k\geq 2\sqrt{10}+\frac{15}{2}\text{ or }k\leq \frac{15}{2}-2\sqrt{10}$$