Whakaoti i te 4sqrt{x-2}=x+1

$$4 \sqrt{ x-2 } =x+1$$

Whakaoti mō x

$$x=11$$

$$x=3$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

$$\left(4\sqrt{x-2}\right)^{2}=\left(x+1\right)^{2}$$

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

$$4^{2}\left(\sqrt{x-2}\right)^{2}=\left(x+1\right)^{2}$$

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

$$16\left(\sqrt{x-2}\right)^{2}=\left(x+1\right)^{2}$$

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

$$16\left(x-2\right)=\left(x+1\right)^{2}$$

Whakamahia te āhuatanga tohatoha hei whakarea te 16 ki te x-2.

$$16x-32=\left(x+1\right)^{2}$$

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

$$16x-32=x^{2}+2x+1$$

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

$$16x-32-x^{2}=2x+1$$

Tangohia te 2x mai i ngā taha e rua.

$$16x-32-x^{2}-2x=1$$

Pahekotia te 16x me -2x, ka 14x.

$$14x-32-x^{2}=1$$

Tangohia te 1 mai i ngā taha e rua.

$$14x-32-x^{2}-1=0$$

Tangohia te 1 i te -32, ka -33.

$$14x-33-x^{2}=0$$

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

$$-x^{2}+14x-33=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei -x^{2}+ax+bx-33. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=14 ab=-\left(-33\right)=33$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 33.

$$1,33 3,11$$

Tātaihia te tapeke mō ia takirua.

$$1+33=34 3+11=14$$

Ko te otinga te takirua ka hoatu i te tapeke 14.

$$a=11 b=3$$

Tuhia anō te -x^{2}+14x-33 hei \left(-x^{2}+11x\right)+\left(3x-33\right).

$$\left(-x^{2}+11x\right)+\left(3x-33\right)$$

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

$$-x\left(x-11\right)+3\left(x-11\right)$$

Whakatauwehea atu te kīanga pātahi x-11 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-11\right)\left(-x+3\right)$$

Hei kimi otinga whārite, me whakaoti te x-11=0 me te -x+3=0.

$$x=11 x=3$$

Whakakapia te 11 mō te x i te whārite 4\sqrt{x-2}=x+1.

$$4\sqrt{11-2}=11+1$$

Whakarūnātia. Ko te uara x=11 kua ngata te whārite.

$$12=12$$

Whakakapia te 3 mō te x i te whārite 4\sqrt{x-2}=x+1.

$$4\sqrt{3-2}=3+1$$

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

$$4=4$$

Rārangihia ngā rongoā katoa o 4\sqrt{x-2}=x+1.

$$x=11 x=3$$