Whakaoti i te sqrt{x+5}=sqrt{8-x}+1

$$\sqrt { x + 5 } = \sqrt { 8 - x } + 1$$

Whakaoti mō x

$$x=4$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

$$\left(\sqrt{x+5}\right)^{2}=\left(\sqrt{8-x}+1\right)^{2}$$

Tātaihia te \sqrt{x+5} mā te pū o 2, kia riro ko x+5.

$$x+5=\left(\sqrt{8-x}+1\right)^{2}$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(\sqrt{8-x}+1\right)^{2}.

$$x+5=\left(\sqrt{8-x}\right)^{2}+2\sqrt{8-x}+1$$

Tātaihia te \sqrt{8-x} mā te pū o 2, kia riro ko 8-x.

$$x+5=8-x+2\sqrt{8-x}+1$$

Tāpirihia te 8 ki te 1, ka 9.

$$x+5=9-x+2\sqrt{8-x}$$

Me tango 9-x mai i ngā taha e rua o te whārite.

$$x+5-\left(9-x\right)=2\sqrt{8-x}$$

Hei kimi i te tauaro o 9-x, kimihia te tauaro o ia taurangi.

$$x+5-9+x=2\sqrt{8-x}$$

Tangohia te 9 i te 5, ka -4.

$$x-4+x=2\sqrt{8-x}$$

Pahekotia te x me x, ka 2x.

$$2x-4=2\sqrt{8-x}$$

Pūruatia ngā taha e rua o te whārite.

$$\left(2x-4\right)^{2}=\left(2\sqrt{8-x}\right)^{2}$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(2x-4\right)^{2}.

$$4x^{2}-16x+16=\left(2\sqrt{8-x}\right)^{2}$$

Whakarohaina te \left(2\sqrt{8-x}\right)^{2}.

$$4x^{2}-16x+16=2^{2}\left(\sqrt{8-x}\right)^{2}$$

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

$$4x^{2}-16x+16=4\left(\sqrt{8-x}\right)^{2}$$

Tātaihia te \sqrt{8-x} mā te pū o 2, kia riro ko 8-x.

$$4x^{2}-16x+16=4\left(8-x\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te 8-x.

$$4x^{2}-16x+16=32-4x$$

Tangohia te 32 mai i ngā taha e rua.

$$4x^{2}-16x+16-32=-4x$$

Tangohia te 32 i te 16, ka -16.

$$4x^{2}-16x-16=-4x$$

Me tāpiri te 4x ki ngā taha e rua.

$$4x^{2}-16x-16+4x=0$$

Pahekotia te -16x me 4x, ka -12x.

$$4x^{2}-12x-16=0$$

Whakawehea ngā taha e rua ki te 4.

$$x^{2}-3x-4=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx-4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-3 ab=1\left(-4\right)=-4$$

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 -4.

$$1,-4 2,-2$$

Tātaihia te tapeke mō ia takirua.

$$1-4=-3 2-2=0$$

Ko te otinga te takirua ka hoatu i te tapeke -3.

$$a=-4 b=1$$

Tuhia anō te x^{2}-3x-4 hei \left(x^{2}-4x\right)+\left(x-4\right).

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

Whakatauwehea atu x i te x^{2}-4x.

$$x\left(x-4\right)+x-4$$

Whakatauwehea atu te kīanga pātahi x-4 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-4\right)\left(x+1\right)$$

Hei kimi otinga whārite, me whakaoti te x-4=0 me te x+1=0.

$$x=4 x=-1$$

Whakakapia te 4 mō te x i te whārite \sqrt{x+5}=\sqrt{8-x}+1.

$$\sqrt{4+5}=\sqrt{8-4}+1$$

Whakarūnātia. Ko te uara x=4 kua ngata te whārite.

$$3=3$$

Whakakapia te -1 mō te x i te whārite \sqrt{x+5}=\sqrt{8-x}+1.

$$\sqrt{-1+5}=\sqrt{8-\left(-1\right)}+1$$

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

$$2=4$$

Whakakapia te 4 mō te x i te whārite \sqrt{x+5}=\sqrt{8-x}+1.

$$\sqrt{4+5}=\sqrt{8-4}+1$$

Whakarūnātia. Ko te uara x=4 kua ngata te whārite.

$$3=3$$

Ko te whārite \sqrt{x+5}=\sqrt{8-x}+1 he rongoā ahurei.

$$x=4$$