Whakaoti i te 2sqrt{4(t-1)}=sqrt{4(2t-1)}

$$2 \sqrt { 4 ( t - 1 ) } = \sqrt { 4 ( 2 t - 1 ) }$$

Whakaoti mō t

$$t = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te t-1.

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

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

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

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

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

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

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

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te 4t-4.

$$16t-16=\left(\sqrt{4\left(2t-1\right)}\right)^{2}$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te 2t-1.

$$16t-16=\left(\sqrt{8t-4}\right)^{2}$$

Tātaihia te \sqrt{8t-4} mā te pū o 2, kia riro ko 8t-4.

$$16t-16=8t-4$$

Tangohia te 8t mai i ngā taha e rua.

$$16t-16-8t=-4$$

Pahekotia te 16t me -8t, ka 8t.

$$8t-16=-4$$

Me tāpiri te 16 ki ngā taha e rua.

$$8t=-4+16$$

Tāpirihia te -4 ki te 16, ka 12.

$$8t=12$$

Whakawehea ngā taha e rua ki te 8.

$$t=\frac{12}{8}$$

Whakahekea te hautanga \frac{12}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$t=\frac{3}{2}$$

Whakakapia te \frac{3}{2} mō te t i te whārite 2\sqrt{4\left(t-1\right)}=\sqrt{4\left(2t-1\right)}.

$$2\sqrt{4\left(\frac{3}{2}-1\right)}=\sqrt{4\left(2\times \frac{3}{2}-1\right)}$$

Whakarūnātia. Ko te uara t=\frac{3}{2} kua ngata te whārite.

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

Ko te whārite 2\sqrt{4\left(t-1\right)}=\sqrt{4\left(2t-1\right)} he rongoā ahurei.

$$t=\frac{3}{2}$$