Whakaoti i te sqrt{4-3x}=sqrt{x+2}

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

Whakaoti mō x

$$x=\frac{1}{2}=0.5$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

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

Tātaihia te \sqrt{4-3x} mā te pū o 2, kia riro ko 4-3x.

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

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

$$4-3x=x+2$$

Tangohia te x mai i ngā taha e rua.

$$4-3x-x=2$$

Pahekotia te -3x me -x, ka -4x.

$$4-4x=2$$

Tangohia te 4 mai i ngā taha e rua.

$$-4x=2-4$$

Tangohia te 4 i te 2, ka -2.

$$-4x=-2$$

Whakawehea ngā taha e rua ki te -4.

$$x=\frac{-2}{-4}$$

Whakahekea te hautanga \frac{-2}{-4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te -2.

$$x=\frac{1}{2}$$

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

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

Whakarūnātia. Ko te uara x=\frac{1}{2} kua ngata te whārite.

$$\frac{1}{2}\times 10^{\frac{1}{2}}=\frac{1}{2}\times 10^{\frac{1}{2}}$$

Ko te whārite \sqrt{4-3x}=\sqrt{x+2} he rongoā ahurei.

$$x=\frac{1}{2}$$