Whakaoti i te (2x-4)(x-4)=(5-x)(4-x)

$$( 2 x - 4 ) ( x - 4 ) = ( 5 - x ) ( 4 - x )$$

Whakaoti mō x

$$x=-1$$

$$x=4$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te 2x-4 ki te x-4 ka whakakotahi i ngā kupu rite.

$$2x^{2}-12x+16=\left(5-x\right)\left(4-x\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 5-x ki te 4-x ka whakakotahi i ngā kupu rite.

$$2x^{2}-12x+16=20-9x+x^{2}$$

Tangohia te 20 mai i ngā taha e rua.

$$2x^{2}-12x+16-20=-9x+x^{2}$$

Tangohia te 20 i te 16, ka -4.

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

Me tāpiri te 9x ki ngā taha e rua.

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

Pahekotia te -12x me 9x, ka -3x.

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

Tangohia te x^{2} mai i ngā taha e rua.

$$2x^{2}-3x-4-x^{2}=0$$

Pahekotia te 2x^{2} me -x^{2}, ka x^{2}.

$$x^{2}-3x-4=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -3 mō b, me -4 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua -3.

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

Whakareatia -4 ki te -4.

$$x=\frac{-\left(-3\right)±\sqrt{9+16}}{2}$$

Tāpiri 9 ki te 16.

$$x=\frac{-\left(-3\right)±\sqrt{25}}{2}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-\left(-3\right)±5}{2}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±5}{2}$$

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

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

Whakawehe 8 ki te 2.

$$x=4$$

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

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

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 2x-4 ki te x-4 ka whakakotahi i ngā kupu rite.

$$2x^{2}-12x+16=\left(5-x\right)\left(4-x\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 5-x ki te 4-x ka whakakotahi i ngā kupu rite.

$$2x^{2}-12x+16=20-9x+x^{2}$$

Me tāpiri te 9x ki ngā taha e rua.

$$2x^{2}-12x+16+9x=20+x^{2}$$

Pahekotia te -12x me 9x, ka -3x.

$$2x^{2}-3x+16=20+x^{2}$$

Tangohia te x^{2} mai i ngā taha e rua.

$$2x^{2}-3x+16-x^{2}=20$$

Pahekotia te 2x^{2} me -x^{2}, ka x^{2}.

$$x^{2}-3x+16=20$$

Tangohia te 16 mai i ngā taha e rua.

$$x^{2}-3x=20-16$$

Tangohia te 16 i te 20, ka 4.

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

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.

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

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

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

Tāpiri 4 ki te \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{25}{4}$$

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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{25}{4}$$

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

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Whakarūnātia.

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

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

$$x=4 x=-1$$