Whakaoti i te -12u-9=4u^2

$$- 12 u - 9 = 4 u ^ { 2 }$$

Whakaoti mō u

$$u = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tangohia te 4u^{2} mai i ngā taha e rua.

$$-12u-9-4u^{2}=0$$

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

$$-4u^{2}-12u-9=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei -4u^{2}+au+bu-9. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-12 ab=-4\left(-9\right)=36$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. 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.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

Ko te otinga te takirua ka hoatu i te tapeke -12.

$$a=-6 b=-6$$

Tuhia anō te -4u^{2}-12u-9 hei \left(-4u^{2}-6u\right)+\left(-6u-9\right).

$$\left(-4u^{2}-6u\right)+\left(-6u-9\right)$$

Tauwehea te 2u i te tuatahi me te 3 i te rōpū tuarua.

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

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

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

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

$$u=-\frac{3}{2} u=-\frac{3}{2}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 4u^{2} mai i ngā taha e rua.

$$-12u-9-4u^{2}=0$$

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.

$$-4u^{2}-12u-9=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -4 mō a, -12 mō b, me -9 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$u=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-4\right)\left(-9\right)}}{2\left(-4\right)}$$

Pūrua -12.

$$u=\frac{-\left(-12\right)±\sqrt{144-4\left(-4\right)\left(-9\right)}}{2\left(-4\right)}$$

Whakareatia -4 ki te -4.

$$u=\frac{-\left(-12\right)±\sqrt{144+16\left(-9\right)}}{2\left(-4\right)}$$

Whakareatia 16 ki te -9.

$$u=\frac{-\left(-12\right)±\sqrt{144-144}}{2\left(-4\right)}$$

Tāpiri 144 ki te -144.

$$u=\frac{-\left(-12\right)±\sqrt{0}}{2\left(-4\right)}$$

Tuhia te pūtakerua o te 0.

$$u=-\frac{-12}{2\left(-4\right)}$$

Ko te tauaro o -12 ko 12.

$$u=\frac{12}{2\left(-4\right)}$$

Whakareatia 2 ki te -4.

$$u=\frac{12}{-8}$$

Whakahekea te hautanga \frac{12}{-8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$u=-\frac{3}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 4u^{2} mai i ngā taha e rua.

$$-12u-9-4u^{2}=0$$

Me tāpiri te 9 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$-12u-4u^{2}=9$$

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.

$$-4u^{2}-12u=9$$

Whakawehea ngā taha e rua ki te -4.

$$\frac{-4u^{2}-12u}{-4}=\frac{9}{-4}$$

Mā te whakawehe ki te -4 ka wetekia te whakareanga ki te -4.

$$u^{2}+\left(-\frac{12}{-4}\right)u=\frac{9}{-4}$$

Whakawehe -12 ki te -4.

$$u^{2}+3u=\frac{9}{-4}$$

Whakawehe 9 ki te -4.

$$u^{2}+3u=-\frac{9}{4}$$

Whakawehea te 3, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{3}{2}. Nā, tāpiria te pūrua o te \frac{3}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$u^{2}+3u+\left(\frac{3}{2}\right)^{2}=-\frac{9}{4}+\left(\frac{3}{2}\right)^{2}$$

Pūruatia \frac{3}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$u^{2}+3u+\frac{9}{4}=\frac{-9+9}{4}$$

Tāpiri -\frac{9}{4} ki te \frac{9}{4} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$u^{2}+3u+\frac{9}{4}=0$$

Tauwehea u^{2}+3u+\frac{9}{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}.

$$\left(u+\frac{3}{2}\right)^{2}=0$$

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

$$\sqrt{\left(u+\frac{3}{2}\right)^{2}}=\sqrt{0}$$

Whakarūnātia.

$$u+\frac{3}{2}=0 u+\frac{3}{2}=0$$

Me tango \frac{3}{2} mai i ngā taha e rua o te whārite.

$$u=-\frac{3}{2} u=-\frac{3}{2}$$

Kua oti te whārite te whakatau. He ōrite ngā whakatau.

$$u=-\frac{3}{2}$$