Réitigh 3x^2+(2k+1)x+3k-2=0

$$3 x ^ { 2 } + ( 2 k + 1 ) x + 3 k - 2 = 0$$

Réitigh do k.

$$k=-\frac{\left(3x-2\right)\left(x+1\right)}{2x+3}$$

$$x\neq -\frac{3}{2}$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

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

$$3x^{2}+2kx+x+3k-2=0$$

Bain 3x^{2} ón dá thaobh. Is ionann rud ar bith a dhealaítear ó nialas agus a shéanadh.

$$2kx+x+3k-2=-3x^{2}$$

Bain x ón dá thaobh.

$$2kx+3k-2=-3x^{2}-x$$

Cuir 2 leis an dá thaobh.

$$2kx+3k=-3x^{2}-x+2$$

Comhcheangail na téarmaí ar fad ina bhfuil k.

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

Tá an chothromóid i bhfoirm chaighdeánach.

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

Roinn an dá thaobh faoi 2x+3.

$$\frac{\left(2x+3\right)k}{2x+3}=-\frac{\left(3x-2\right)\left(x+1\right)}{2x+3}$$

Má roinntear é faoi 2x+3 cuirtear an iolrúchán faoi 2x+3 ar ceal.

$$k=-\frac{\left(3x-2\right)\left(x+1\right)}{2x+3}$$

Réitigh do x. (complex solution)

$$x=\frac{\sqrt{4k^{2}-32k+25}}{6}-\frac{k}{3}-\frac{1}{6}$$

$$x=-\frac{\sqrt{4k^{2}-32k+25}}{6}-\frac{k}{3}-\frac{1}{6}$$

Réitigh do x.

$$x=\frac{\sqrt{4k^{2}-32k+25}}{6}-\frac{k}{3}-\frac{1}{6}$$

$$x=-\frac{\sqrt{4k^{2}-32k+25}}{6}-\frac{k}{3}-\frac{1}{6}\text{, }k\geq \frac{\sqrt{39}}{2}+4\text{ or }k\leq -\frac{\sqrt{39}}{2}+4$$