Whakaoti i te left(1--1/2right)x^2+x+1-k=0

$$\left( 1- \frac{ -1 }{ 2 } \right) { x }^{ 2 } +x+1-k = 0$$

Whakaoti mō k

$$k=\frac{3x^{2}}{2}+x+1$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Ka taea te hautanga \frac{-1}{2} te tuhi anō ko -\frac{1}{2} mā te tango i te tohu tōraro.

$$\left(1-\left(-\frac{1}{2}\right)\right)x^{2}+x+1-k=0$$

Ko te tauaro o -\frac{1}{2} ko \frac{1}{2}.

$$\left(1+\frac{1}{2}\right)x^{2}+x+1-k=0$$

Tāpirihia te 1 ki te \frac{1}{2}, ka \frac{3}{2}.

$$\frac{3}{2}x^{2}+x+1-k=0$$

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

$$x+1-k=-\frac{3}{2}x^{2}$$

Tangohia te x mai i ngā taha e rua.

$$1-k=-\frac{3}{2}x^{2}-x$$

Tangohia te 1 mai i ngā taha e rua.

$$-k=-\frac{3}{2}x^{2}-x-1$$

He hanga arowhānui tō te whārite.

$$-k=-\frac{3x^{2}}{2}-x-1$$

Whakawehea ngā taha e rua ki te -1.

$$\frac{-k}{-1}=\frac{-\frac{3x^{2}}{2}-x-1}{-1}$$

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

$$k=\frac{-\frac{3x^{2}}{2}-x-1}{-1}$$

Whakawehe -\frac{3x^{2}}{2}-x-1 ki te -1.

$$k=\frac{3x^{2}}{2}+x+1$$

Whakaoti mō x (complex solution)

$$x=\frac{\sqrt{6k-5}-1}{3}$$

$$x=\frac{-\sqrt{6k-5}-1}{3}$$

Whakaoti mō x

$$x=\frac{\sqrt{6k-5}-1}{3}$$

$$x=\frac{-\sqrt{6k-5}-1}{3}\text{, }k\geq \frac{5}{6}$$