Whakaoti i te 3q^2-12q=15

$$3 q ^ { 2 } - 12 q = 15$$

Whakaoti mō q

$$q=-1$$

$$q=5$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tangohia te 15 mai i ngā taha e rua.

$$3q^{2}-12q-15=0$$

Whakawehea ngā taha e rua ki te 3.

$$q^{2}-4q-5=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei q^{2}+aq+bq-5. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-4 ab=1\left(-5\right)=-5$$

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. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-5 b=1$$

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

$$\left(q^{2}-5q\right)+\left(q-5\right)$$

Whakatauwehea atu q i te q^{2}-5q.

$$q\left(q-5\right)+q-5$$

Whakatauwehea atu te kīanga pātahi q-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(q-5\right)\left(q+1\right)$$

Hei kimi otinga whārite, me whakaoti te q-5=0 me te q+1=0.

$$q=5 q=-1$$

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.

$$3q^{2}-12q=15$$

Me tango 15 mai i ngā taha e rua o te whārite.

$$3q^{2}-12q-15=15-15$$

Mā te tango i te 15 i a ia ake anō ka toe ko te 0.

$$3q^{2}-12q-15=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, -12 mō b, me -15 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 3\left(-15\right)}}{2\times 3}$$

Pūrua -12.

$$q=\frac{-\left(-12\right)±\sqrt{144-4\times 3\left(-15\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$q=\frac{-\left(-12\right)±\sqrt{144-12\left(-15\right)}}{2\times 3}$$

Whakareatia -12 ki te -15.

$$q=\frac{-\left(-12\right)±\sqrt{144+180}}{2\times 3}$$

Tāpiri 144 ki te 180.

$$q=\frac{-\left(-12\right)±\sqrt{324}}{2\times 3}$$

Tuhia te pūtakerua o te 324.

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

Ko te tauaro o -12 ko 12.

$$q=\frac{12±18}{2\times 3}$$

Whakareatia 2 ki te 3.

$$q=\frac{12±18}{6}$$

Nā, me whakaoti te whārite q=\frac{12±18}{6} ina he tāpiri te ±. Tāpiri 12 ki te 18.

$$q=\frac{30}{6}$$

Whakawehe 30 ki te 6.

$$q=5$$

Nā, me whakaoti te whārite q=\frac{12±18}{6} ina he tango te ±. Tango 18 mai i 12.

$$q=-\frac{6}{6}$$

Whakawehe -6 ki te 6.

$$q=-1$$

Kua oti te whārite te whakatau.

$$q=5 q=-1$$

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.

$$3q^{2}-12q=15$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{3q^{2}-12q}{3}=\frac{15}{3}$$

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

$$q^{2}+\left(-\frac{12}{3}\right)q=\frac{15}{3}$$

Whakawehe -12 ki te 3.

$$q^{2}-4q=\frac{15}{3}$$

Whakawehe 15 ki te 3.

$$q^{2}-4q=5$$

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

$$q^{2}-4q+\left(-2\right)^{2}=5+\left(-2\right)^{2}$$

Pūrua -2.

$$q^{2}-4q+4=5+4$$

Tāpiri 5 ki te 4.

$$q^{2}-4q+4=9$$

Tauwehea q^{2}-4q+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}.

$$\left(q-2\right)^{2}=9$$

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

$$\sqrt{\left(q-2\right)^{2}}=\sqrt{9}$$

Whakarūnātia.

$$q-2=3 q-2=-3$$

Me tāpiri 2 ki ngā taha e rua o te whārite.

$$q=5 q=-1$$