Whakaoti i te sqrt{q+2}+1=sqrt{3q+7}

$$\sqrt { q + 2 } + 1 = \sqrt { 3 q + 7 }$$

Whakaoti mō q

$$q=-1$$

$$q=-2$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

$$\left(\sqrt{q+2}+1\right)^{2}=\left(\sqrt{3q+7}\right)^{2}$$

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

$$\left(\sqrt{q+2}\right)^{2}+2\sqrt{q+2}+1=\left(\sqrt{3q+7}\right)^{2}$$

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

$$q+2+2\sqrt{q+2}+1=\left(\sqrt{3q+7}\right)^{2}$$

Tāpirihia te 2 ki te 1, ka 3.

$$q+3+2\sqrt{q+2}=\left(\sqrt{3q+7}\right)^{2}$$

Tātaihia te \sqrt{3q+7} mā te pū o 2, kia riro ko 3q+7.

$$q+3+2\sqrt{q+2}=3q+7$$

Me tango q+3 mai i ngā taha e rua o te whārite.

$$2\sqrt{q+2}=3q+7-\left(q+3\right)$$

Hei kimi i te tauaro o q+3, kimihia te tauaro o ia taurangi.

$$2\sqrt{q+2}=3q+7-q-3$$

Pahekotia te 3q me -q, ka 2q.

$$2\sqrt{q+2}=2q+7-3$$

Tangohia te 3 i te 7, ka 4.

$$2\sqrt{q+2}=2q+4$$

Pūruatia ngā taha e rua o te whārite.

$$\left(2\sqrt{q+2}\right)^{2}=\left(2q+4\right)^{2}$$

Whakarohaina te \left(2\sqrt{q+2}\right)^{2}.

$$2^{2}\left(\sqrt{q+2}\right)^{2}=\left(2q+4\right)^{2}$$

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

$$4\left(\sqrt{q+2}\right)^{2}=\left(2q+4\right)^{2}$$

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

$$4\left(q+2\right)=\left(2q+4\right)^{2}$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te q+2.

$$4q+8=\left(2q+4\right)^{2}$$

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

$$4q+8=4q^{2}+16q+16$$

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

$$4q+8-4q^{2}=16q+16$$

Tangohia te 16q mai i ngā taha e rua.

$$4q+8-4q^{2}-16q=16$$

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

$$-12q+8-4q^{2}=16$$

Tangohia te 16 mai i ngā taha e rua.

$$-12q+8-4q^{2}-16=0$$

Tangohia te 16 i te 8, ka -8.

$$-12q-8-4q^{2}=0$$

Whakawehea ngā taha e rua ki te 4.

$$-3q-2-q^{2}=0$$

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

$$-q^{2}-3q-2=0$$

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

$$a+b=-3 ab=-\left(-2\right)=2$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-1 b=-2$$

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

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

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

$$q\left(-q-1\right)+2\left(-q-1\right)$$

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

$$\left(-q-1\right)\left(q+2\right)$$

Hei kimi otinga whārite, me whakaoti te -q-1=0 me te q+2=0.

$$q=-1 q=-2$$

Whakakapia te -1 mō te q i te whārite \sqrt{q+2}+1=\sqrt{3q+7}.

$$\sqrt{-1+2}+1=\sqrt{3\left(-1\right)+7}$$

Whakarūnātia. Ko te uara q=-1 kua ngata te whārite.

$$2=2$$

Whakakapia te -2 mō te q i te whārite \sqrt{q+2}+1=\sqrt{3q+7}.

$$\sqrt{-2+2}+1=\sqrt{3\left(-2\right)+7}$$

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

$$1=1$$

Rārangihia ngā rongoā katoa o \sqrt{q+2}+1=\sqrt{3q+7}.

$$q=-1 q=-2$$