Whakaoti i te 3x^2+(2k+1)x+3k-2=0

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

Whakaoti mō k

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

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

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te 2k+1 ki te x.

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

Tangohia te 3x^{2} mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

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

Tangohia te x mai i ngā taha e rua.

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

Me tāpiri te 2 ki ngā taha e rua.

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

Pahekotia ngā kīanga tau katoa e whai ana i te k.

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

He hanga arowhānui tō te whārite.

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

Whakawehea ngā taha e rua ki te 2x+3.

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

Mā te whakawehe ki te 2x+3 ka wetekia te whakareanga ki te 2x+3.

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

Whakaoti mō x (complex solution)

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

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

Whakaoti mō x

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

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