Whakaoti i te sqrt{x+3}+sqrt{x+6}=sqrt{x+11}

$$\sqrt { x + 3 } + \sqrt { x + 6 } = \sqrt { x + 11 }$$

Whakaoti mō x

$$x=-2$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

$$\left(\sqrt{x+3}+\sqrt{x+6}\right)^{2}=\left(\sqrt{x+11}\right)^{2}$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(\sqrt{x+3}+\sqrt{x+6}\right)^{2}.

$$\left(\sqrt{x+3}\right)^{2}+2\sqrt{x+3}\sqrt{x+6}+\left(\sqrt{x+6}\right)^{2}=\left(\sqrt{x+11}\right)^{2}$$

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

$$x+3+2\sqrt{x+3}\sqrt{x+6}+\left(\sqrt{x+6}\right)^{2}=\left(\sqrt{x+11}\right)^{2}$$

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

$$x+3+2\sqrt{x+3}\sqrt{x+6}+x+6=\left(\sqrt{x+11}\right)^{2}$$

Pahekotia te x me x, ka 2x.

$$2x+3+2\sqrt{x+3}\sqrt{x+6}+6=\left(\sqrt{x+11}\right)^{2}$$

Tāpirihia te 3 ki te 6, ka 9.

$$2x+9+2\sqrt{x+3}\sqrt{x+6}=\left(\sqrt{x+11}\right)^{2}$$

Tātaihia te \sqrt{x+11} mā te pū o 2, kia riro ko x+11.

$$2x+9+2\sqrt{x+3}\sqrt{x+6}=x+11$$

Me tango 2x+9 mai i ngā taha e rua o te whārite.

$$2\sqrt{x+3}\sqrt{x+6}=x+11-\left(2x+9\right)$$

Hei kimi i te tauaro o 2x+9, kimihia te tauaro o ia taurangi.

$$2\sqrt{x+3}\sqrt{x+6}=x+11-2x-9$$

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

$$2\sqrt{x+3}\sqrt{x+6}=-x+11-9$$

Tangohia te 9 i te 11, ka 2.

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

Pūruatia ngā taha e rua o te whārite.

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

Whakarohaina te \left(2\sqrt{x+3}\sqrt{x+6}\right)^{2}.

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

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

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

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

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

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

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

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te x+3.

$$\left(4x+12\right)\left(x+6\right)=\left(-x+2\right)^{2}$$

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o 4x+12 ki ia tau o x+6.

$$4x^{2}+24x+12x+72=\left(-x+2\right)^{2}$$

Pahekotia te 24x me 12x, ka 36x.

$$4x^{2}+36x+72=\left(-x+2\right)^{2}$$

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

$$4x^{2}+36x+72=x^{2}-4x+4$$

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

$$4x^{2}+36x+72-x^{2}=-4x+4$$

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

$$3x^{2}+36x+72=-4x+4$$

Me tāpiri te 4x ki ngā taha e rua.

$$3x^{2}+36x+72+4x=4$$

Pahekotia te 36x me 4x, ka 40x.

$$3x^{2}+40x+72=4$$

Tangohia te 4 mai i ngā taha e rua.

$$3x^{2}+40x+72-4=0$$

Tangohia te 4 i te 72, ka 68.

$$3x^{2}+40x+68=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 3x^{2}+ax+bx+68. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=40 ab=3\times 68=204$$

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 204.

$$1,204 2,102 3,68 4,51 6,34 12,17$$

Tātaihia te tapeke mō ia takirua.

$$1+204=205 2+102=104 3+68=71 4+51=55 6+34=40 12+17=29$$

Ko te otinga te takirua ka hoatu i te tapeke 40.

$$a=6 b=34$$

Tuhia anō te 3x^{2}+40x+68 hei \left(3x^{2}+6x\right)+\left(34x+68\right).

$$\left(3x^{2}+6x\right)+\left(34x+68\right)$$

Tauwehea te 3x i te tuatahi me te 34 i te rōpū tuarua.

$$3x\left(x+2\right)+34\left(x+2\right)$$

Whakatauwehea atu te kīanga pātahi x+2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x+2\right)\left(3x+34\right)$$

Hei kimi otinga whārite, me whakaoti te x+2=0 me te 3x+34=0.

$$x=-2 x=-\frac{34}{3}$$

Whakakapia te -\frac{34}{3} mō te x i te whārite \sqrt{x+3}+\sqrt{x+6}=\sqrt{x+11}. Te kīanga \sqrt{-\frac{34}{3}+3} kia kore e tautuhitia nā te mea kāore te radicand e noho tōraro.

$$\sqrt{-\frac{34}{3}+3}+\sqrt{-\frac{34}{3}+6}=\sqrt{-\frac{34}{3}+11}$$

Whakakapia te -2 mō te x i te whārite \sqrt{x+3}+\sqrt{x+6}=\sqrt{x+11}.

$$\sqrt{-2+3}+\sqrt{-2+6}=\sqrt{-2+11}$$

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

$$3=3$$

Ko te whārite \sqrt{x+3}+\sqrt{x+6}=\sqrt{x+11} he rongoā ahurei.

$$x=-2$$