Whakaoti i te {l}{-2x=0}{text{Solvefor}ytext{where}}{y={(sqrt{3}-1)}^2+{(2x-2)}{(sqrt{3}-1)}+2}

$$\left. \begin{array} { l } { -2 x = 0 }\\ { \text{Solve for } y \text{ where} } \\ { y = {(\sqrt{3} - 1)} ^ {2} + {(2 x - 2)} {(\sqrt{3} - 1)} + 2 } \end{array} \right.$$

Whakaoti mō x, y

$$y=8-4\sqrt{3}\approx 1.07179677$$

Ngā Upane Otinga

Whakaarohia te whārite tuatahi. Whakawehea ngā taha e rua ki te -2. Ko te kore i whakawehea ki te tau ehara te kore ka hua ko te kore.

$$x=0$$

Whakaarohia te whārite tuarua. Me kōkuhu ngā uara tāupe mōhiotia ki te whārite.

$$y=\left(\sqrt{3}-1\right)^{2}+\left(2\times 0-2\right)\left(\sqrt{3}-1\right)+2$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(\sqrt{3}-1\right)^{2}.

$$y=\left(\sqrt{3}\right)^{2}-2\sqrt{3}+1+\left(2\times 0-2\right)\left(\sqrt{3}-1\right)+2$$

Ko te pūrua o \sqrt{3} ko 3.

$$y=3-2\sqrt{3}+1+\left(2\times 0-2\right)\left(\sqrt{3}-1\right)+2$$

Tāpirihia te 3 ki te 1, ka 4.

$$y=4-2\sqrt{3}+\left(2\times 0-2\right)\left(\sqrt{3}-1\right)+2$$

Whakareatia te 2 ki te 0, ka 0.

$$y=4-2\sqrt{3}+\left(0-2\right)\left(\sqrt{3}-1\right)+2$$

Tangohia te 2 i te 0, ka -2.

$$y=4-2\sqrt{3}-2\left(\sqrt{3}-1\right)+2$$

Whakamahia te āhuatanga tohatoha hei whakarea te -2 ki te \sqrt{3}-1.

$$y=4-2\sqrt{3}-2\sqrt{3}+2+2$$

Pahekotia te -2\sqrt{3} me -2\sqrt{3}, ka -4\sqrt{3}.

$$y=4-4\sqrt{3}+2+2$$

Tāpirihia te 4 ki te 2, ka 6.

$$y=6-4\sqrt{3}+2$$

Tāpirihia te 6 ki te 2, ka 8.

$$y=8-4\sqrt{3}$$

Kua oti te pūnaha te whakatau.

$$x=0 y=8-4\sqrt{3}$$