Whakaoti i te 2sqrt{x+5}=x+2?

$$2 \sqrt { x + 5 } = x + 2 ?$$

Whakaoti mō x

$$x=4$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

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

Whakarohaina te \left(2\sqrt{x+5}\right)^{2}.

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

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

$$4\left(\sqrt{x+5}\right)^{2}=\left(x+2\right)^{2}$$

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

$$4\left(x+5\right)=\left(x+2\right)^{2}$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te x+5.

$$4x+20=\left(x+2\right)^{2}$$

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

$$4x+20=x^{2}+4x+4$$

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

$$4x+20-x^{2}=4x+4$$

Tangohia te 4x mai i ngā taha e rua.

$$4x+20-x^{2}-4x=4$$

Pahekotia te 4x me -4x, ka 0.

$$20-x^{2}=4$$

Tangohia te 20 mai i ngā taha e rua.

$$-x^{2}=4-20$$

Tangohia te 20 i te 4, ka -16.

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

Whakawehea ngā taha e rua ki te -1.

$$x^{2}=\frac{-16}{-1}$$

Ka taea te hautanga \frac{-16}{-1} te whakamāmā ki te 16 mā te tango tahi i te tohu tōraro i te taurunga me te tauraro.

$$x^{2}=16$$

Tuhia te pūtakerua o ngā taha e rua o te whārite.

$$x=4 x=-4$$

Whakakapia te 4 mō te x i te whārite 2\sqrt{x+5}=x+2.

$$2\sqrt{4+5}=4+2$$

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

$$6=6$$

Whakakapia te -4 mō te x i te whārite 2\sqrt{x+5}=x+2.

$$2\sqrt{-4+5}=-4+2$$

Whakarūnātia. Ko te uara x=-4 kāore e ngata ana ki te whārite nā te mea e rerekē ngā tohu o te taha maui me te taha katau.

$$2=-2$$

Ko te whārite 2\sqrt{x+5}=x+2 he rongoā ahurei.

$$x=4$$