Whakaoti i te (x-1)(x+2)+3x=4(x-2)-(x-12)

$$( x - 1 ) ( x + 2 ) + 3 x = 4 ( x - 2 ) - ( x - 12 )$$

Whakaoti mō x

$$x=-3$$

$$x=2$$

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.

$$x^{2}+x-2+3x=4\left(x-2\right)-\left(x-12\right)$$

Pahekotia te x me 3x, ka 4x.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te x-2.

$$x^{2}+4x-2=4x-8-\left(x-12\right)$$

Hei kimi i te tauaro o x-12, kimihia te tauaro o ia taurangi.

$$x^{2}+4x-2=4x-8-x+12$$

Pahekotia te 4x me -x, ka 3x.

$$x^{2}+4x-2=3x-8+12$$

Tāpirihia te -8 ki te 12, ka 4.

$$x^{2}+4x-2=3x+4$$

Tangohia te 3x mai i ngā taha e rua.

$$x^{2}+4x-2-3x=4$$

Pahekotia te 4x me -3x, ka x.

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

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}.

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

Pūrua 1.

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

Whakareatia -4 ki te -6.

$$x=\frac{-1±\sqrt{1+24}}{2}$$

Tāpiri 1 ki te 24.

$$x=\frac{-1±\sqrt{25}}{2}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-1±5}{2}$$

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

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

Whakawehe 4 ki te 2.

$$x=2$$

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

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

Whakawehe -6 ki te 2.

$$x=-3$$

Kua oti te whārite te whakatau.

$$x=2 x=-3$$

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.

$$x^{2}+x-2+3x=4\left(x-2\right)-\left(x-12\right)$$

Pahekotia te x me 3x, ka 4x.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te x-2.

$$x^{2}+4x-2=4x-8-\left(x-12\right)$$

Hei kimi i te tauaro o x-12, kimihia te tauaro o ia taurangi.

$$x^{2}+4x-2=4x-8-x+12$$

Pahekotia te 4x me -x, ka 3x.

$$x^{2}+4x-2=3x-8+12$$

Tāpirihia te -8 ki te 12, ka 4.

$$x^{2}+4x-2=3x+4$$

Tangohia te 3x mai i ngā taha e rua.

$$x^{2}+4x-2-3x=4$$

Pahekotia te 4x me -3x, ka x.

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

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

$$x=2 x=-3$$