Whakaoti i te 2x^2-5x-18=0

$$2 { x }^{ 2 } -5x-18 = 0$$

Whakaoti mō x

$$x=-2$$

$$x = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 2x^{2}+ax+bx-18. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-5 ab=2\left(-18\right)=-36$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Tātaihia te tapeke mō ia takirua.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

Ko te otinga te takirua ka hoatu i te tapeke -5.

$$a=-9 b=4$$

Tuhia anō te 2x^{2}-5x-18 hei \left(2x^{2}-9x\right)+\left(4x-18\right).

$$\left(2x^{2}-9x\right)+\left(4x-18\right)$$

Tauwehea te x i te tuatahi me te 2 i te rōpū tuarua.

$$x\left(2x-9\right)+2\left(2x-9\right)$$

Whakatauwehea atu te kīanga pātahi 2x-9 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2x-9\right)\left(x+2\right)$$

Hei kimi otinga whārite, me whakaoti te 2x-9=0 me te x+2=0.

$$x=\frac{9}{2} x=-2$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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.

$$2x^{2}-5x-18=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, -5 mō b, me -18 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2\left(-18\right)}}{2\times 2}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\left(-18\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\left(-18\right)}}{2\times 2}$$

Whakareatia -8 ki te -18.

$$x=\frac{-\left(-5\right)±\sqrt{25+144}}{2\times 2}$$

Tāpiri 25 ki te 144.

$$x=\frac{-\left(-5\right)±\sqrt{169}}{2\times 2}$$

Tuhia te pūtakerua o te 169.

$$x=\frac{-\left(-5\right)±13}{2\times 2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±13}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{5±13}{4}$$

Nā, me whakaoti te whārite x=\frac{5±13}{4} ina he tāpiri te ±. Tāpiri 5 ki te 13.

$$x=\frac{18}{4}$$

Whakahekea te hautanga \frac{18}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{9}{2}$$

Nā, me whakaoti te whārite x=\frac{5±13}{4} ina he tango te ±. Tango 13 mai i 5.

$$x=-\frac{8}{4}$$

Whakawehe -8 ki te 4.

$$x=-2$$

Kua oti te whārite te whakatau.

$$x=\frac{9}{2} x=-2$$

Ngā Upane mō te Whakaoti i te Pūrua

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.

$$2x^{2}-5x-18=0$$

Me tāpiri 18 ki ngā taha e rua o te whārite.

$$2x^{2}-5x-18-\left(-18\right)=-\left(-18\right)$$

Mā te tango i te -18 i a ia ake anō ka toe ko te 0.

$$2x^{2}-5x=-\left(-18\right)$$

Tango -18 mai i 0.

$$2x^{2}-5x=18$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}-5x}{2}=\frac{18}{2}$$

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

$$x^{2}-\frac{5}{2}x=\frac{18}{2}$$

Whakawehe 18 ki te 2.

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

Whakawehea te -\frac{5}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{5}{4}. Nā, tāpiria te pūrua o te -\frac{5}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia -\frac{5}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=9+\frac{25}{16}$$

Tāpiri 9 ki te \frac{25}{16}.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{169}{16}$$

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

$$\left(x-\frac{5}{4}\right)^{2}=\frac{169}{16}$$

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

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Whakarūnātia.

$$x-\frac{5}{4}=\frac{13}{4} x-\frac{5}{4}=-\frac{13}{4}$$

Me tāpiri \frac{5}{4} ki ngā taha e rua o te whārite.

$$x=\frac{9}{2} x=-2$$