Whakaoti i te (6-x)(2x-5)+30=0

$$(6-x)(2x-5)+30=0$$

Whakaoti mō x

$$x = \frac{17}{2} = 8\frac{1}{2} = 8.5$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Whakamahia te āhuatanga tuaritanga hei whakarea te 6-x ki te 2x-5 ka whakakotahi i ngā kupu rite.

$$17x-30-2x^{2}+30=0$$

Tāpirihia te -30 ki te 30, ka 0.

$$17x-2x^{2}=0$$

Tauwehea te x.

$$x\left(17-2x\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 17-2x=0.

$$x=0 x=\frac{17}{2}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te 6-x ki te 2x-5 ka whakakotahi i ngā kupu rite.

$$17x-30-2x^{2}+30=0$$

Tāpirihia te -30 ki te 30, ka 0.

$$17x-2x^{2}=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$-2x^{2}+17x=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -2 mō a, 17 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-17±\sqrt{17^{2}}}{2\left(-2\right)}$$

Tuhia te pūtakerua o te 17^{2}.

$$x=\frac{-17±17}{2\left(-2\right)}$$

Whakareatia 2 ki te -2.

$$x=\frac{-17±17}{-4}$$

Nā, me whakaoti te whārite x=\frac{-17±17}{-4} ina he tāpiri te ±. Tāpiri -17 ki te 17.

$$x=\frac{0}{-4}$$

Whakawehe 0 ki te -4.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-17±17}{-4} ina he tango te ±. Tango 17 mai i -17.

$$x=-\frac{34}{-4}$$

Whakahekea te hautanga \frac{-34}{-4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{17}{2}$$

Kua oti te whārite te whakatau.

$$x=0 x=\frac{17}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te 6-x ki te 2x-5 ka whakakotahi i ngā kupu rite.

$$17x-30-2x^{2}+30=0$$

Tāpirihia te -30 ki te 30, ka 0.

$$17x-2x^{2}=0$$

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

$$-2x^{2}+17x=0$$

Whakawehea ngā taha e rua ki te -2.

$$\frac{-2x^{2}+17x}{-2}=\frac{0}{-2}$$

Mā te whakawehe ki te -2 ka wetekia te whakareanga ki te -2.

$$x^{2}+\frac{17}{-2}x=\frac{0}{-2}$$

Whakawehe 17 ki te -2.

$$x^{2}-\frac{17}{2}x=\frac{0}{-2}$$

Whakawehe 0 ki te -2.

$$x^{2}-\frac{17}{2}x=0$$

Whakawehea te -\frac{17}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{17}{4}. Nā, tāpiria te pūrua o te -\frac{17}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{17}{2}x+\left(-\frac{17}{4}\right)^{2}=\left(-\frac{17}{4}\right)^{2}$$

Pūruatia -\frac{17}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{17}{2}x+\frac{289}{16}=\frac{289}{16}$$

Tauwehea x^{2}-\frac{17}{2}x+\frac{289}{16}. 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}.

$$\left(x-\frac{17}{4}\right)^{2}=\frac{289}{16}$$

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

$$\sqrt{\left(x-\frac{17}{4}\right)^{2}}=\sqrt{\frac{289}{16}}$$

Whakarūnātia.

$$x-\frac{17}{4}=\frac{17}{4} x-\frac{17}{4}=-\frac{17}{4}$$

Me tāpiri \frac{17}{4} ki ngā taha e rua o te whārite.

$$x=\frac{17}{2} x=0$$