Whakaoti i te 4x(x+1)=15

$$4x(x+1)=15$$

Whakaoti mō x

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

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 4x ki te x+1.

$$4x^{2}+4x=15$$

Tangohia te 15 mai i ngā taha e rua.

$$4x^{2}+4x-15=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 4 mō a, 4 mō b, me -15 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 4\left(-15\right)}}{2\times 4}$$

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\times 4\left(-15\right)}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-4±\sqrt{16-16\left(-15\right)}}{2\times 4}$$

Whakareatia -16 ki te -15.

$$x=\frac{-4±\sqrt{16+240}}{2\times 4}$$

Tāpiri 16 ki te 240.

$$x=\frac{-4±\sqrt{256}}{2\times 4}$$

Tuhia te pūtakerua o te 256.

$$x=\frac{-4±16}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-4±16}{8}$$

Nā, me whakaoti te whārite x=\frac{-4±16}{8} ina he tāpiri te ±. Tāpiri -4 ki te 16.

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

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

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

Nā, me whakaoti te whārite x=\frac{-4±16}{8} ina he tango te ±. Tango 16 mai i -4.

$$x=-\frac{20}{8}$$

Whakahekea te hautanga \frac{-20}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$x=-\frac{5}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{3}{2} x=-\frac{5}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 4x ki te x+1.

$$4x^{2}+4x=15$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4x^{2}+4x}{4}=\frac{15}{4}$$

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

$$x^{2}+\frac{4}{4}x=\frac{15}{4}$$

Whakawehe 4 ki te 4.

$$x^{2}+x=\frac{15}{4}$$

Whakawehea te 1, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{2}. Nā, tāpiria te pūrua o te \frac{1}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{15}{4}+\left(\frac{1}{2}\right)^{2}$$

Pūruatia \frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+x+\frac{1}{4}=\frac{15+1}{4}$$

Tāpiri \frac{15}{4} ki te \frac{1}{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.

$$x^{2}+x+\frac{1}{4}=4$$

Tauwehea x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=4$$

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

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{4}$$

Whakarūnātia.

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

Me tango \frac{1}{2} mai i ngā taha e rua o te whārite.

$$x=\frac{3}{2} x=-\frac{5}{2}$$