Whakaoti i te 2x^2+14x-4=-x^2+3x

$$2 x ^ { 2 } + 14 x - 4 = - x ^ { 2 } + 3 x$$

Whakaoti mō x

$$x=-4$$

$$x=\frac{1}{3}\approx 0.333333333$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Me tāpiri te x^{2} ki ngā taha e rua.

$$2x^{2}+14x-4+x^{2}=3x$$

Pahekotia te 2x^{2} me x^{2}, ka 3x^{2}.

$$3x^{2}+14x-4=3x$$

Tangohia te 3x mai i ngā taha e rua.

$$3x^{2}+14x-4-3x=0$$

Pahekotia te 14x me -3x, ka 11x.

$$3x^{2}+11x-4=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 3x^{2}+ax+bx-4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=11 ab=3\left(-4\right)=-12$$

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 -12.

$$-1,12 -2,6 -3,4$$

Tātaihia te tapeke mō ia takirua.

$$-1+12=11 -2+6=4 -3+4=1$$

Ko te otinga te takirua ka hoatu i te tapeke 11.

$$a=-1 b=12$$

Tuhia anō te 3x^{2}+11x-4 hei \left(3x^{2}-x\right)+\left(12x-4\right).

$$\left(3x^{2}-x\right)+\left(12x-4\right)$$

Tauwehea te x i te tuatahi me te 4 i te rōpū tuarua.

$$x\left(3x-1\right)+4\left(3x-1\right)$$

Whakatauwehea atu te kīanga pātahi 3x-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3x-1\right)\left(x+4\right)$$

Hei kimi otinga whārite, me whakaoti te 3x-1=0 me te x+4=0.

$$x=\frac{1}{3} x=-4$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Me tāpiri te x^{2} ki ngā taha e rua.

$$2x^{2}+14x-4+x^{2}=3x$$

Pahekotia te 2x^{2} me x^{2}, ka 3x^{2}.

$$3x^{2}+14x-4=3x$$

Tangohia te 3x mai i ngā taha e rua.

$$3x^{2}+14x-4-3x=0$$

Pahekotia te 14x me -3x, ka 11x.

$$3x^{2}+11x-4=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, 11 mō b, me -4 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\times 3\left(-4\right)}}{2\times 3}$$

Pūrua 11.

$$x=\frac{-11±\sqrt{121-4\times 3\left(-4\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-11±\sqrt{121-12\left(-4\right)}}{2\times 3}$$

Whakareatia -12 ki te -4.

$$x=\frac{-11±\sqrt{121+48}}{2\times 3}$$

Tāpiri 121 ki te 48.

$$x=\frac{-11±\sqrt{169}}{2\times 3}$$

Tuhia te pūtakerua o te 169.

$$x=\frac{-11±13}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-11±13}{6}$$

Nā, me whakaoti te whārite x=\frac{-11±13}{6} ina he tāpiri te ±. Tāpiri -11 ki te 13.

$$x=\frac{2}{6}$$

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

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

Nā, me whakaoti te whārite x=\frac{-11±13}{6} ina he tango te ±. Tango 13 mai i -11.

$$x=-\frac{24}{6}$$

Whakawehe -24 ki te 6.

$$x=-4$$

Kua oti te whārite te whakatau.

$$x=\frac{1}{3} x=-4$$

Ngā Upane mō te Whakaoti i te Pūrua

Me tāpiri te x^{2} ki ngā taha e rua.

$$2x^{2}+14x-4+x^{2}=3x$$

Pahekotia te 2x^{2} me x^{2}, ka 3x^{2}.

$$3x^{2}+14x-4=3x$$

Tangohia te 3x mai i ngā taha e rua.

$$3x^{2}+14x-4-3x=0$$

Pahekotia te 14x me -3x, ka 11x.

$$3x^{2}+11x-4=0$$

Me tāpiri te 4 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$3x^{2}+11x=4$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{3x^{2}+11x}{3}=\frac{4}{3}$$

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

$$x^{2}+\frac{11}{3}x=\frac{4}{3}$$

Whakawehea te \frac{11}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{11}{6}. Nā, tāpiria te pūrua o te \frac{11}{6} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{11}{3}x+\left(\frac{11}{6}\right)^{2}=\frac{4}{3}+\left(\frac{11}{6}\right)^{2}$$

Pūruatia \frac{11}{6} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{11}{3}x+\frac{121}{36}=\frac{4}{3}+\frac{121}{36}$$

Tāpiri \frac{4}{3} ki te \frac{121}{36} 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.

$$x^{2}+\frac{11}{3}x+\frac{121}{36}=\frac{169}{36}$$

Tauwehea x^{2}+\frac{11}{3}x+\frac{121}{36}. 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}.

$$\left(x+\frac{11}{6}\right)^{2}=\frac{169}{36}$$

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

$$\sqrt{\left(x+\frac{11}{6}\right)^{2}}=\sqrt{\frac{169}{36}}$$

Whakarūnātia.

$$x+\frac{11}{6}=\frac{13}{6} x+\frac{11}{6}=-\frac{13}{6}$$

Me tango \frac{11}{6} mai i ngā taha e rua o te whārite.

$$x=\frac{1}{3} x=-4$$