Whakaoti i te (2x-1)*(2x+1)=12x-10

$$(2x-1) \times (2x+1)=12x-10$$

Whakaoti mō x

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakaarohia te \left(2x-1\right)\left(2x+1\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Pūrua 1.

$$\left(2x\right)^{2}-1=12x-10$$

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

$$2^{2}x^{2}-1=12x-10$$

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

$$4x^{2}-1=12x-10$$

Tangohia te 12x mai i ngā taha e rua.

$$4x^{2}-1-12x=-10$$

Me tāpiri te 10 ki ngā taha e rua.

$$4x^{2}-1-12x+10=0$$

Tāpirihia te -1 ki te 10, ka 9.

$$4x^{2}+9-12x=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$4x^{2}-12x+9=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 4 mō a, -12 mō b, me 9 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\times 9}}{2\times 4}$$

Pūrua -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 4\times 9}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-\left(-12\right)±\sqrt{144-16\times 9}}{2\times 4}$$

Whakareatia -16 ki te 9.

$$x=\frac{-\left(-12\right)±\sqrt{144-144}}{2\times 4}$$

Tāpiri 144 ki te -144.

$$x=\frac{-\left(-12\right)±\sqrt{0}}{2\times 4}$$

Tuhia te pūtakerua o te 0.

$$x=-\frac{-12}{2\times 4}$$

Ko te tauaro o -12 ko 12.

$$x=\frac{12}{2\times 4}$$

Whakareatia 2 ki te 4.

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

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

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

Ngā Upane mō te Whakaoti i te Pūrua

Whakaarohia te \left(2x-1\right)\left(2x+1\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Pūrua 1.

$$\left(2x\right)^{2}-1=12x-10$$

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

$$2^{2}x^{2}-1=12x-10$$

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

$$4x^{2}-1=12x-10$$

Tangohia te 12x mai i ngā taha e rua.

$$4x^{2}-1-12x=-10$$

Me tāpiri te 1 ki ngā taha e rua.

$$4x^{2}-12x=-10+1$$

Tāpirihia te -10 ki te 1, ka -9.

$$4x^{2}-12x=-9$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4x^{2}-12x}{4}=-\frac{9}{4}$$

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

$$x^{2}+\left(-\frac{12}{4}\right)x=-\frac{9}{4}$$

Whakawehe -12 ki te 4.

$$x^{2}-3x=-\frac{9}{4}$$

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

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

Pūruatia -\frac{3}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-3x+\frac{9}{4}=\frac{-9+9}{4}$$

Tāpiri -\frac{9}{4} ki te \frac{9}{4} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x^{2}-3x+\frac{9}{4}=0$$

Tauwehea x^{2}-3x+\frac{9}{4}. 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}.

$$\left(x-\frac{3}{2}\right)^{2}=0$$

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

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{0}$$

Whakarūnātia.

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

Me tāpiri \frac{3}{2} ki ngā taha e rua o te whārite.

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

Kua oti te whārite te whakatau. He ōrite ngā whakatau.

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