Whakaoti i te sqrt{x^2-2x+10}+2x+1=0

$$\sqrt{ { x }^{ 2 } -2x+10 } +2x+1 = 0$$

Whakaoti mō x

$$x=-3$$

Ngā Upane Otinga

Me tango 2x+1 mai i ngā taha e rua o te whārite.

$$\sqrt{x^{2}-2x+10}=-\left(2x+1\right)$$

Hei kimi i te tauaro o 2x+1, kimihia te tauaro o ia taurangi.

$$\sqrt{x^{2}-2x+10}=-2x-1$$

Pūruatia ngā taha e rua o te whārite.

$$\left(\sqrt{x^{2}-2x+10}\right)^{2}=\left(-2x-1\right)^{2}$$

Tātaihia te \sqrt{x^{2}-2x+10} mā te pū o 2, kia riro ko x^{2}-2x+10.

$$x^{2}-2x+10=\left(-2x-1\right)^{2}$$

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

$$x^{2}-2x+10=4x^{2}+4x+1$$

Tangohia te 4x^{2} mai i ngā taha e rua.

$$x^{2}-2x+10-4x^{2}=4x+1$$

Pahekotia te x^{2} me -4x^{2}, ka -3x^{2}.

$$-3x^{2}-2x+10=4x+1$$

Tangohia te 4x mai i ngā taha e rua.

$$-3x^{2}-2x+10-4x=1$$

Pahekotia te -2x me -4x, ka -6x.

$$-3x^{2}-6x+10=1$$

Tangohia te 1 mai i ngā taha e rua.

$$-3x^{2}-6x+10-1=0$$

Tangohia te 1 i te 10, ka 9.

$$-3x^{2}-6x+9=0$$

Whakawehea ngā taha e rua ki te 3.

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

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei -x^{2}+ax+bx+3. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-2 ab=-3=-3$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=1 b=-3$$

Tuhia anō te -x^{2}-2x+3 hei \left(-x^{2}+x\right)+\left(-3x+3\right).

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

Tauwehea te x i te tuatahi me te 3 i te rōpū tuarua.

$$x\left(-x+1\right)+3\left(-x+1\right)$$

Whakatauwehea atu te kīanga pātahi -x+1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(-x+1\right)\left(x+3\right)$$

Hei kimi otinga whārite, me whakaoti te -x+1=0 me te x+3=0.

$$x=1 x=-3$$

Whakakapia te 1 mō te x i te whārite \sqrt{x^{2}-2x+10}+2x+1=0.

$$\sqrt{1^{2}-2+10}+2\times 1+1=0$$

Whakarūnātia. Ko te uara x=1 kāore e ngata ana ki te whārite.

$$6=0$$

Whakakapia te -3 mō te x i te whārite \sqrt{x^{2}-2x+10}+2x+1=0.

$$\sqrt{\left(-3\right)^{2}-2\left(-3\right)+10}+2\left(-3\right)+1=0$$

Whakarūnātia. Ko te uara x=-3 kua ngata te whārite.

$$0=0$$

Ko te whārite \sqrt{x^{2}-2x+10}=-2x-1 he rongoā ahurei.

$$x=-3$$