Whakaoti i te left(1-9/4right)x^2+x+1-k=0

$$\left( 1- \frac{ 9 }{ 4 } \right) { x }^{ 2 } +x+1-k = 0$$

Whakaoti mō k

$$k=-\frac{5x^{2}}{4}+x+1$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tangohia te \frac{9}{4} i te 1, ka -\frac{5}{4}.

$$-\frac{5}{4}x^{2}+x+1-k=0$$

Me tāpiri te \frac{5}{4}x^{2} ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$x+1-k=\frac{5}{4}x^{2}$$

Tangohia te x mai i ngā taha e rua.

$$1-k=\frac{5}{4}x^{2}-x$$

Tangohia te 1 mai i ngā taha e rua.

$$-k=\frac{5}{4}x^{2}-x-1$$

He hanga arowhānui tō te whārite.

$$-k=\frac{5x^{2}}{4}-x-1$$

Whakawehea ngā taha e rua ki te -1.

$$\frac{-k}{-1}=\frac{\frac{5x^{2}}{4}-x-1}{-1}$$

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

$$k=\frac{\frac{5x^{2}}{4}-x-1}{-1}$$

Whakawehe \frac{5x^{2}}{4}-x-1 ki te -1.

$$k=-\frac{5x^{2}}{4}+x+1$$

Whakaoti mō x (complex solution)

$$x=\frac{-2\sqrt{6-5k}+2}{5}$$

$$x=\frac{2\sqrt{6-5k}+2}{5}$$

Whakaoti mō x

$$x=\frac{-2\sqrt{6-5k}+2}{5}$$

$$x=\frac{2\sqrt{6-5k}+2}{5}\text{, }k\leq \frac{6}{5}$$