Whakaoti i te 100x^2-100x-600=0

$$100 x ^ { 2 } - 100 x - 600 = 0$$

Whakaoti mō x

$$x=-2$$

$$x=3$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakawehea ngā taha e rua ki te 100.

$$x^{2}-x-6=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx-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 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 -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.

$$a=-3 b=2$$

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

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

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

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

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

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

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

$$x=3 x=-2$$

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.

$$100x^{2}-100x-600=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 100 mō a, -100 mō b, me -600 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-100\right)±\sqrt{\left(-100\right)^{2}-4\times 100\left(-600\right)}}{2\times 100}$$

Pūrua -100.

$$x=\frac{-\left(-100\right)±\sqrt{10000-4\times 100\left(-600\right)}}{2\times 100}$$

Whakareatia -4 ki te 100.

$$x=\frac{-\left(-100\right)±\sqrt{10000-400\left(-600\right)}}{2\times 100}$$

Whakareatia -400 ki te -600.

$$x=\frac{-\left(-100\right)±\sqrt{10000+240000}}{2\times 100}$$

Tāpiri 10000 ki te 240000.

$$x=\frac{-\left(-100\right)±\sqrt{250000}}{2\times 100}$$

Tuhia te pūtakerua o te 250000.

$$x=\frac{-\left(-100\right)±500}{2\times 100}$$

Ko te tauaro o -100 ko 100.

$$x=\frac{100±500}{2\times 100}$$

Whakareatia 2 ki te 100.

$$x=\frac{100±500}{200}$$

Nā, me whakaoti te whārite x=\frac{100±500}{200} ina he tāpiri te ±. Tāpiri 100 ki te 500.

$$x=\frac{600}{200}$$

Whakawehe 600 ki te 200.

$$x=3$$

Nā, me whakaoti te whārite x=\frac{100±500}{200} ina he tango te ±. Tango 500 mai i 100.

$$x=-\frac{400}{200}$$

Whakawehe -400 ki te 200.

$$x=-2$$

Kua oti te whārite te whakatau.

$$x=3 x=-2$$

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.

$$100x^{2}-100x-600=0$$

Me tāpiri 600 ki ngā taha e rua o te whārite.

$$100x^{2}-100x-600-\left(-600\right)=-\left(-600\right)$$

Mā te tango i te -600 i a ia ake anō ka toe ko te 0.

$$100x^{2}-100x=-\left(-600\right)$$

Tango -600 mai i 0.

$$100x^{2}-100x=600$$

Whakawehea ngā taha e rua ki te 100.

$$\frac{100x^{2}-100x}{100}=\frac{600}{100}$$

Mā te whakawehe ki te 100 ka wetekia te whakareanga ki te 100.

$$x^{2}+\left(-\frac{100}{100}\right)x=\frac{600}{100}$$

Whakawehe -100 ki te 100.

$$x^{2}-x=\frac{600}{100}$$

Whakawehe 600 ki te 100.

$$x^{2}-x=6$$

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.

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

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

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

Tāpiri 6 ki te \frac{1}{4}.

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

Tauwehea x^{2}-x+\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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{25}{4}$$

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

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

Whakarūnātia.

$$x-\frac{1}{2}=\frac{5}{2} x-\frac{1}{2}=-\frac{5}{2}$$

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

$$x=3 x=-2$$