Whakaoti i te (x+1)*(x-2)=4

$$(x+1) \times (x-2)=4$$

Whakaoti mō x

$$x=-2$$

$$x=3$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te x+1 ki te x-2 ka whakakotahi i ngā kupu rite.

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

Tangohia te 4 mai i ngā taha e rua.

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

Tangohia te 4 i te -2, ka -6.

$$x^{2}-x-6=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -1 mō b, me -6 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-6\right)}}{2}$$

Whakareatia -4 ki te -6.

$$x=\frac{-\left(-1\right)±\sqrt{1+24}}{2}$$

Tāpiri 1 ki te 24.

$$x=\frac{-\left(-1\right)±\sqrt{25}}{2}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-\left(-1\right)±5}{2}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±5}{2}$$

Nā, me whakaoti te whārite x=\frac{1±5}{2} ina he tāpiri te ±. Tāpiri 1 ki te 5.

$$x=\frac{6}{2}$$

Whakawehe 6 ki te 2.

$$x=3$$

Nā, me whakaoti te whārite x=\frac{1±5}{2} ina he tango te ±. Tango 5 mai i 1.

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

Whakawehe -4 ki te 2.

$$x=-2$$

Kua oti te whārite te whakatau.

$$x=3 x=-2$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te x+1 ki te x-2 ka whakakotahi i ngā kupu rite.

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

Me tāpiri te 2 ki ngā taha e rua.

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

Tāpirihia te 4 ki te 2, ka 6.

$$x^{2}-x=6$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=6+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=6+\frac{1}{4}$$

Tāpiri 6 ki te \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{25}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{25}{4}$$

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

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Whakarūnātia.

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

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

$$x=3 x=-2$$