Whakaoti i te x+1=sqrt{(3x+7)}

$$x + 1 = \sqrt { ( 3 x + 7 ) }$$

Whakaoti mō x

$$x=3$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

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

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

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

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

$$x^{2}+2x+1=3x+7$$

Tangohia te 3x mai i ngā taha e rua.

$$x^{2}+2x+1-3x=7$$

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

$$x^{2}-x+1=7$$

Tangohia te 7 mai i ngā taha e rua.

$$x^{2}-x+1-7=0$$

Tangohia te 7 i te 1, ka -6.

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

Hei whakaoti i te whārite, whakatauwehea te x^{2}-x-6 mā te whakamahi i te tātai x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-1 ab=-6$$

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.

$$a=-3 b=2$$

Me tuhi anō te kīanga whakatauwehe \left(x+a\right)\left(x+b\right) mā ngā uara i tātaihia.

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

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 x+1=\sqrt{3x+7}.

$$3+1=\sqrt{3\times 3+7}$$

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

$$4=4$$

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

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

Whakarūnātia. Ko te uara x=-2 kāore e ngata ana ki te whārite nā te mea e rerekē ngā tohu o te taha maui me te taha katau.

$$-1=1$$

Ko te whārite x+1=\sqrt{3x+7} he rongoā ahurei.

$$x=3$$