Whakaoti i te x_1^2-2-(x-0.8)=-2(0.9+x)

$$x _ { 1 } ^ { 2 } - 2 - ( x - 0.8 ) = - 2 ( 0.9 + x )$$

Whakaoti mō x

$$x=-\left(x_{1}^{2}+0.6\right)$$

Ngā Upane Otinga

Hei kimi i te tauaro o x-0.8, kimihia te tauaro o ia taurangi.

$$x_{1}^{2}-2-x+0.8=-2\left(0.9+x\right)$$

Tāpirihia te -2 ki te 0.8, ka -1.2.

$$x_{1}^{2}-1.2-x=-2\left(0.9+x\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te -2 ki te 0.9+x.

$$x_{1}^{2}-1.2-x=-1.8-2x$$

Me tāpiri te 2x ki ngā taha e rua.

$$x_{1}^{2}-1.2-x+2x=-1.8$$

Pahekotia te -x me 2x, ka x.

$$x_{1}^{2}-1.2+x=-1.8$$

Tangohia te x_{1}^{2} mai i ngā taha e rua.

$$-1.2+x=-1.8-x_{1}^{2}$$

Me tāpiri te 1.2 ki ngā taha e rua.

$$x=-1.8-x_{1}^{2}+1.2$$

Tāpirihia te -1.8 ki te 1.2, ka -0.6.

$$x=-0.6-x_{1}^{2}$$

Whakaoti mō x_1 (complex solution)

$$x_{1}=-i\sqrt{x+0.6}$$

$$x_{1}=i\sqrt{x+0.6}$$

Whakaoti mō x_1

$$x_{1}=\frac{\sqrt{-4x-2.4}}{2}$$

$$x_{1}=-\frac{\sqrt{-4x-2.4}}{2}\text{, }x\leq -\frac{3}{5}$$