Whakaoti i te 2x^2+13x-24=0

$$2 { x }^{ 2 } +13x-24=0$$

Whakaoti mō x

$$x=-8$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

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-24. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=13 ab=2\left(-24\right)=-48$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -48.

$$-1,48 -2,24 -3,16 -4,12 -6,8$$

Tātaihia te tapeke mō ia takirua.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

Ko te otinga te takirua ka hoatu i te tapeke 13.

$$a=-3 b=16$$

Tuhia anō te 2x^{2}+13x-24 hei \left(2x^{2}-3x\right)+\left(16x-24\right).

$$\left(2x^{2}-3x\right)+\left(16x-24\right)$$

Tauwehea te x i te tuatahi me te 8 i te rōpū tuarua.

$$x\left(2x-3\right)+8\left(2x-3\right)$$

Whakatauwehea atu te kīanga pātahi 2x-3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2x-3\right)\left(x+8\right)$$

Hei kimi otinga whārite, me whakaoti te 2x-3=0 me te x+8=0.

$$x=\frac{3}{2} x=-8$$

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.

$$2x^{2}+13x-24=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, 13 mō b, me -24 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-13±\sqrt{13^{2}-4\times 2\left(-24\right)}}{2\times 2}$$

Pūrua 13.

$$x=\frac{-13±\sqrt{169-4\times 2\left(-24\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-13±\sqrt{169-8\left(-24\right)}}{2\times 2}$$

Whakareatia -8 ki te -24.

$$x=\frac{-13±\sqrt{169+192}}{2\times 2}$$

Tāpiri 169 ki te 192.

$$x=\frac{-13±\sqrt{361}}{2\times 2}$$

Tuhia te pūtakerua o te 361.

$$x=\frac{-13±19}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-13±19}{4}$$

Nā, me whakaoti te whārite x=\frac{-13±19}{4} ina he tāpiri te ±. Tāpiri -13 ki te 19.

$$x=\frac{6}{4}$$

Whakahekea te hautanga \frac{6}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Nā, me whakaoti te whārite x=\frac{-13±19}{4} ina he tango te ±. Tango 19 mai i -13.

$$x=-\frac{32}{4}$$

Whakawehe -32 ki te 4.

$$x=-8$$

Kua oti te whārite te whakatau.

$$x=\frac{3}{2} x=-8$$

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.

$$2x^{2}+13x-24=0$$

Me tāpiri 24 ki ngā taha e rua o te whārite.

$$2x^{2}+13x-24-\left(-24\right)=-\left(-24\right)$$

Mā te tango i te -24 i a ia ake anō ka toe ko te 0.

$$2x^{2}+13x=-\left(-24\right)$$

Tango -24 mai i 0.

$$2x^{2}+13x=24$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}+13x}{2}=\frac{24}{2}$$

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

$$x^{2}+\frac{13}{2}x=\frac{24}{2}$$

Whakawehe 24 ki te 2.

$$x^{2}+\frac{13}{2}x=12$$

Whakawehea te \frac{13}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{13}{4}. Nā, tāpiria te pūrua o te \frac{13}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{13}{2}x+\left(\frac{13}{4}\right)^{2}=12+\left(\frac{13}{4}\right)^{2}$$

Pūruatia \frac{13}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{13}{2}x+\frac{169}{16}=12+\frac{169}{16}$$

Tāpiri 12 ki te \frac{169}{16}.

$$x^{2}+\frac{13}{2}x+\frac{169}{16}=\frac{361}{16}$$

Tauwehea x^{2}+\frac{13}{2}x+\frac{169}{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}.

$$\left(x+\frac{13}{4}\right)^{2}=\frac{361}{16}$$

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

$$\sqrt{\left(x+\frac{13}{4}\right)^{2}}=\sqrt{\frac{361}{16}}$$

Whakarūnātia.

$$x+\frac{13}{4}=\frac{19}{4} x+\frac{13}{4}=-\frac{19}{4}$$

Me tango \frac{13}{4} mai i ngā taha e rua o te whārite.

$$x=\frac{3}{2} x=-8$$