Whakaoti i te sqrt{10-3x}=2+sqrt{x+6}

$$\sqrt { 10 - 3 x } = 2 + \sqrt { x + 6 }$$

Whakaoti mō x

$$x=-2$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

$$\left(\sqrt{10-3x}\right)^{2}=\left(2+\sqrt{x+6}\right)^{2}$$

Tātaihia te \sqrt{10-3x} mā te pū o 2, kia riro ko 10-3x.

$$10-3x=\left(2+\sqrt{x+6}\right)^{2}$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(2+\sqrt{x+6}\right)^{2}.

$$10-3x=4+4\sqrt{x+6}+\left(\sqrt{x+6}\right)^{2}$$

Tātaihia te \sqrt{x+6} mā te pū o 2, kia riro ko x+6.

$$10-3x=4+4\sqrt{x+6}+x+6$$

Tāpirihia te 4 ki te 6, ka 10.

$$10-3x=10+4\sqrt{x+6}+x$$

Me tango 10+x mai i ngā taha e rua o te whārite.

$$10-3x-\left(10+x\right)=4\sqrt{x+6}$$

Hei kimi i te tauaro o 10+x, kimihia te tauaro o ia taurangi.

$$10-3x-10-x=4\sqrt{x+6}$$

Tangohia te 10 i te 10, ka 0.

$$-3x-x=4\sqrt{x+6}$$

Pahekotia te -3x me -x, ka -4x.

$$-4x=4\sqrt{x+6}$$

Pūruatia ngā taha e rua o te whārite.

$$\left(-4x\right)^{2}=\left(4\sqrt{x+6}\right)^{2}$$

Whakarohaina te \left(-4x\right)^{2}.

$$\left(-4\right)^{2}x^{2}=\left(4\sqrt{x+6}\right)^{2}$$

Tātaihia te -4 mā te pū o 2, kia riro ko 16.

$$16x^{2}=\left(4\sqrt{x+6}\right)^{2}$$

Whakarohaina te \left(4\sqrt{x+6}\right)^{2}.

$$16x^{2}=4^{2}\left(\sqrt{x+6}\right)^{2}$$

Tātaihia te 4 mā te pū o 2, kia riro ko 16.

$$16x^{2}=16\left(\sqrt{x+6}\right)^{2}$$

Tātaihia te \sqrt{x+6} mā te pū o 2, kia riro ko x+6.

$$16x^{2}=16\left(x+6\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 16 ki te x+6.

$$16x^{2}=16x+96$$

Tangohia te 16x mai i ngā taha e rua.

$$16x^{2}-16x=96$$

Tangohia te 96 mai i ngā taha e rua.

$$16x^{2}-16x-96=0$$

Whakawehea ngā taha e rua ki te 16.

$$x^{2}-x-6=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx-6. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-1 ab=1\left(-6\right)=-6$$

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. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -6.

$$1,-6 2,-3$$

Tātaihia te tapeke mō ia takirua.

$$1-6=-5 2-3=-1$$

Ko te otinga te takirua ka hoatu i te tapeke -1.

$$a=-3 b=2$$

Tuhia anō te x^{2}-x-6 hei \left(x^{2}-3x\right)+\left(2x-6\right).

$$\left(x^{2}-3x\right)+\left(2x-6\right)$$

Tauwehea te x i te tuatahi me te 2 i te rōpū tuarua.

$$x\left(x-3\right)+2\left(x-3\right)$$

Whakatauwehea atu te kīanga pātahi x-3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-3\right)\left(x+2\right)$$

Hei kimi otinga whārite, me whakaoti te x-3=0 me te x+2=0.

$$x=3 x=-2$$

Whakakapia te 3 mō te x i te whārite \sqrt{10-3x}=2+\sqrt{x+6}.

$$\sqrt{10-3\times 3}=2+\sqrt{3+6}$$

Whakarūnātia. Ko te uara x=3 kāore e ngata ana ki te whārite.

$$1=5$$

Whakakapia te -2 mō te x i te whārite \sqrt{10-3x}=2+\sqrt{x+6}.

$$\sqrt{10-3\left(-2\right)}=2+\sqrt{-2+6}$$

Whakarūnātia. Ko te uara x=-2 kua ngata te whārite.

$$4=4$$

Ko te whārite \sqrt{10-3x}=\sqrt{x+6}+2 he rongoā ahurei.

$$x=-2$$