Whakaoti i te 3f^2+3f-18=0

$$3 f ^ { 2 } + 3 f - 18 = 0$$

Whakaoti mō f

$$f=-3$$

$$f=2$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakawehea ngā taha e rua ki te 3.

$$f^{2}+f-6=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei f^{2}+af+bf-6. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=1 ab=1\left(-6\right)=-6$$

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 -6.

$$-1,6 -2,3$$

Tātaihia te tapeke mō ia takirua.

$$-1+6=5 -2+3=1$$

Ko te otinga te takirua ka hoatu i te tapeke 1.

$$a=-2 b=3$$

Tuhia anō te f^{2}+f-6 hei \left(f^{2}-2f\right)+\left(3f-6\right).

$$\left(f^{2}-2f\right)+\left(3f-6\right)$$

Tauwehea te f i te tuatahi me te 3 i te rōpū tuarua.

$$f\left(f-2\right)+3\left(f-2\right)$$

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

$$\left(f-2\right)\left(f+3\right)$$

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

$$f=2 f=-3$$

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.

$$3f^{2}+3f-18=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, 3 mō b, me -18 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$f=\frac{-3±\sqrt{3^{2}-4\times 3\left(-18\right)}}{2\times 3}$$

Pūrua 3.

$$f=\frac{-3±\sqrt{9-4\times 3\left(-18\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$f=\frac{-3±\sqrt{9-12\left(-18\right)}}{2\times 3}$$

Whakareatia -12 ki te -18.

$$f=\frac{-3±\sqrt{9+216}}{2\times 3}$$

Tāpiri 9 ki te 216.

$$f=\frac{-3±\sqrt{225}}{2\times 3}$$

Tuhia te pūtakerua o te 225.

$$f=\frac{-3±15}{2\times 3}$$

Whakareatia 2 ki te 3.

$$f=\frac{-3±15}{6}$$

Nā, me whakaoti te whārite f=\frac{-3±15}{6} ina he tāpiri te ±. Tāpiri -3 ki te 15.

$$f=\frac{12}{6}$$

Whakawehe 12 ki te 6.

$$f=2$$

Nā, me whakaoti te whārite f=\frac{-3±15}{6} ina he tango te ±. Tango 15 mai i -3.

$$f=-\frac{18}{6}$$

Whakawehe -18 ki te 6.

$$f=-3$$

Kua oti te whārite te whakatau.

$$f=2 f=-3$$

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.

$$3f^{2}+3f-18=0$$

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

$$3f^{2}+3f-18-\left(-18\right)=-\left(-18\right)$$

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

$$3f^{2}+3f=-\left(-18\right)$$

Tango -18 mai i 0.

$$3f^{2}+3f=18$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{3f^{2}+3f}{3}=\frac{18}{3}$$

Mā te whakawehe ki te 3 ka wetekia te whakareanga ki te 3.

$$f^{2}+\frac{3}{3}f=\frac{18}{3}$$

Whakawehe 3 ki te 3.

$$f^{2}+f=\frac{18}{3}$$

Whakawehe 18 ki te 3.

$$f^{2}+f=6$$

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.

$$f^{2}+f+\left(\frac{1}{2}\right)^{2}=6+\left(\frac{1}{2}\right)^{2}$$

Pūruatia \frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$f^{2}+f+\frac{1}{4}=6+\frac{1}{4}$$

Tāpiri 6 ki te \frac{1}{4}.

$$f^{2}+f+\frac{1}{4}=\frac{25}{4}$$

Tauwehea f^{2}+f+\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}.

$$\left(f+\frac{1}{2}\right)^{2}=\frac{25}{4}$$

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

$$\sqrt{\left(f+\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Whakarūnātia.

$$f+\frac{1}{2}=\frac{5}{2} f+\frac{1}{2}=-\frac{5}{2}$$

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

$$f=2 f=-3$$