Whakaoti i te (2t+3)^2=3(2t+3)

$$( 2 t + 3 ) ^ { 2 } = 3 ( 2 t + 3 )$$

Whakaoti mō t

$$t=0$$

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

Ngā Upane Whakamahi Tauwehe

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

$$4t^{2}+12t+9=3\left(2t+3\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3 ki te 2t+3.

$$4t^{2}+12t+9=6t+9$$

Tangohia te 6t mai i ngā taha e rua.

$$4t^{2}+12t+9-6t=9$$

Pahekotia te 12t me -6t, ka 6t.

$$4t^{2}+6t+9=9$$

Tangohia te 9 mai i ngā taha e rua.

$$4t^{2}+6t+9-9=0$$

Tangohia te 9 i te 9, ka 0.

$$4t^{2}+6t=0$$

Tauwehea te t.

$$t\left(4t+6\right)=0$$

Hei kimi otinga whārite, me whakaoti te t=0 me te 4t+6=0.

$$t=0 t=-\frac{3}{2}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

$$4t^{2}+12t+9=3\left(2t+3\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3 ki te 2t+3.

$$4t^{2}+12t+9=6t+9$$

Tangohia te 6t mai i ngā taha e rua.

$$4t^{2}+12t+9-6t=9$$

Pahekotia te 12t me -6t, ka 6t.

$$4t^{2}+6t+9=9$$

Tangohia te 9 mai i ngā taha e rua.

$$4t^{2}+6t+9-9=0$$

Tangohia te 9 i te 9, ka 0.

$$4t^{2}+6t=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 4 mō a, 6 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-6±\sqrt{6^{2}}}{2\times 4}$$

Tuhia te pūtakerua o te 6^{2}.

$$t=\frac{-6±6}{2\times 4}$$

Whakareatia 2 ki te 4.

$$t=\frac{-6±6}{8}$$

Nā, me whakaoti te whārite t=\frac{-6±6}{8} ina he tāpiri te ±. Tāpiri -6 ki te 6.

$$t=\frac{0}{8}$$

Whakawehe 0 ki te 8.

$$t=0$$

Nā, me whakaoti te whārite t=\frac{-6±6}{8} ina he tango te ±. Tango 6 mai i -6.

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

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

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

Kua oti te whārite te whakatau.

$$t=0 t=-\frac{3}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

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

$$4t^{2}+12t+9=3\left(2t+3\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3 ki te 2t+3.

$$4t^{2}+12t+9=6t+9$$

Tangohia te 6t mai i ngā taha e rua.

$$4t^{2}+12t+9-6t=9$$

Pahekotia te 12t me -6t, ka 6t.

$$4t^{2}+6t+9=9$$

Tangohia te 9 mai i ngā taha e rua.

$$4t^{2}+6t=9-9$$

Tangohia te 9 i te 9, ka 0.

$$4t^{2}+6t=0$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4t^{2}+6t}{4}=\frac{0}{4}$$

Mā te whakawehe ki te 4 ka wetekia te whakareanga ki te 4.

$$t^{2}+\frac{6}{4}t=\frac{0}{4}$$

Whakahekea te hautanga \frac{6}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$t^{2}+\frac{3}{2}t=\frac{0}{4}$$

Whakawehe 0 ki te 4.

$$t^{2}+\frac{3}{2}t=0$$

Whakawehea te \frac{3}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{3}{4}. Nā, tāpiria te pūrua o te \frac{3}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$t^{2}+\frac{3}{2}t+\left(\frac{3}{4}\right)^{2}=\left(\frac{3}{4}\right)^{2}$$

Pūruatia \frac{3}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$t^{2}+\frac{3}{2}t+\frac{9}{16}=\frac{9}{16}$$

Tauwehea t^{2}+\frac{3}{2}t+\frac{9}{16}. Ko te tikanga pūnoa, ina ko x^{2}+bx+c he pūrua tika pūrua tika pū, ka taea taua mea te tauwehea i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+\frac{3}{4}\right)^{2}=\frac{9}{16}$$

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

$$\sqrt{\left(t+\frac{3}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Whakarūnātia.

$$t+\frac{3}{4}=\frac{3}{4} t+\frac{3}{4}=-\frac{3}{4}$$

Me tango \frac{3}{4} mai i ngā taha e rua o te whārite.

$$t=0 t=-\frac{3}{2}$$