Whakaoti i te (x-3)(x-2)=2

$$( x - 3 ) ( x - 2 ) = 2$$

Whakaoti mō x

$$x=1$$

$$x=4$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te x-3 ki te x-2 ka whakakotahi i ngā kupu rite.

$$x^{2}-5x+6=2$$

Tangohia te 2 mai i ngā taha e rua.

$$x^{2}-5x+6-2=0$$

Tangohia te 2 i te 6, ka 4.

$$x^{2}-5x+4=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -5 mō b, me 4 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 4}}{2}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 4}}{2}$$

Whakareatia -4 ki te 4.

$$x=\frac{-\left(-5\right)±\sqrt{25-16}}{2}$$

Tāpiri 25 ki te -16.

$$x=\frac{-\left(-5\right)±\sqrt{9}}{2}$$

Tuhia te pūtakerua o te 9.

$$x=\frac{-\left(-5\right)±3}{2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±3}{2}$$

Nā, me whakaoti te whārite x=\frac{5±3}{2} ina he tāpiri te ±. Tāpiri 5 ki te 3.

$$x=\frac{8}{2}$$

Whakawehe 8 ki te 2.

$$x=4$$

Nā, me whakaoti te whārite x=\frac{5±3}{2} ina he tango te ±. Tango 3 mai i 5.

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

Whakawehe 2 ki te 2.

$$x=1$$

Kua oti te whārite te whakatau.

$$x=4 x=1$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te x-3 ki te x-2 ka whakakotahi i ngā kupu rite.

$$x^{2}-5x+6=2$$

Tangohia te 6 mai i ngā taha e rua.

$$x^{2}-5x=2-6$$

Tangohia te 6 i te 2, ka -4.

$$x^{2}-5x=-4$$

Whakawehea te -5, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{5}{2}. Nā, tāpiria te pūrua o te -\frac{5}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-4+\left(-\frac{5}{2}\right)^{2}$$

Pūruatia -\frac{5}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-5x+\frac{25}{4}=-4+\frac{25}{4}$$

Tāpiri -4 ki te \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{9}{4}$$

Tauwehea x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{9}{4}$$

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

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Whakarūnātia.

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

Me tāpiri \frac{5}{2} ki ngā taha e rua o te whārite.

$$x=4 x=1$$