Whakaoti i te (4x-3)(x^2+x+3)=(2x+1)(2x^2-3x)

$$( 4 x - 3 ) ( x ^ { 2 } + x + 3 ) = ( 2 x + 1 ) ( 2 x ^ { 2 } - 3 x )$$

Whakaoti mō x

$$x=-3$$

$$x=\frac{3}{5}=0.6$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakamahia te āhuatanga tuaritanga hei whakarea te 4x-3 ki te x^{2}+x+3 ka whakakotahi i ngā kupu rite.

$$4x^{3}+x^{2}+9x-9=\left(2x+1\right)\left(2x^{2}-3x\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 2x+1 ki te 2x^{2}-3x ka whakakotahi i ngā kupu rite.

$$4x^{3}+x^{2}+9x-9=4x^{3}-4x^{2}-3x$$

Tangohia te 4x^{3} mai i ngā taha e rua.

$$4x^{3}+x^{2}+9x-9-4x^{3}=-4x^{2}-3x$$

Pahekotia te 4x^{3} me -4x^{3}, ka 0.

$$x^{2}+9x-9=-4x^{2}-3x$$

Me tāpiri te 4x^{2} ki ngā taha e rua.

$$x^{2}+9x-9+4x^{2}=-3x$$

Pahekotia te x^{2} me 4x^{2}, ka 5x^{2}.

$$5x^{2}+9x-9=-3x$$

Me tāpiri te 3x ki ngā taha e rua.

$$5x^{2}+9x-9+3x=0$$

Pahekotia te 9x me 3x, ka 12x.

$$5x^{2}+12x-9=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 5x^{2}+ax+bx-9. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=12 ab=5\left(-9\right)=-45$$

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 -45.

$$-1,45 -3,15 -5,9$$

Tātaihia te tapeke mō ia takirua.

$$-1+45=44 -3+15=12 -5+9=4$$

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

$$a=-3 b=15$$

Tuhia anō te 5x^{2}+12x-9 hei \left(5x^{2}-3x\right)+\left(15x-9\right).

$$\left(5x^{2}-3x\right)+\left(15x-9\right)$$

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

$$x\left(5x-3\right)+3\left(5x-3\right)$$

Whakatauwehea atu te kīanga pātahi 5x-3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(5x-3\right)\left(x+3\right)$$

Hei kimi otinga whārite, me whakaoti te 5x-3=0 me te x+3=0.

$$x=\frac{3}{5} x=-3$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te 4x-3 ki te x^{2}+x+3 ka whakakotahi i ngā kupu rite.

$$4x^{3}+x^{2}+9x-9=\left(2x+1\right)\left(2x^{2}-3x\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 2x+1 ki te 2x^{2}-3x ka whakakotahi i ngā kupu rite.

$$4x^{3}+x^{2}+9x-9=4x^{3}-4x^{2}-3x$$

Tangohia te 4x^{3} mai i ngā taha e rua.

$$4x^{3}+x^{2}+9x-9-4x^{3}=-4x^{2}-3x$$

Pahekotia te 4x^{3} me -4x^{3}, ka 0.

$$x^{2}+9x-9=-4x^{2}-3x$$

Me tāpiri te 4x^{2} ki ngā taha e rua.

$$x^{2}+9x-9+4x^{2}=-3x$$

Pahekotia te x^{2} me 4x^{2}, ka 5x^{2}.

$$5x^{2}+9x-9=-3x$$

Me tāpiri te 3x ki ngā taha e rua.

$$5x^{2}+9x-9+3x=0$$

Pahekotia te 9x me 3x, ka 12x.

$$5x^{2}+12x-9=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 5 mō a, 12 mō b, me -9 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 5\left(-9\right)}}{2\times 5}$$

Pūrua 12.

$$x=\frac{-12±\sqrt{144-4\times 5\left(-9\right)}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-12±\sqrt{144-20\left(-9\right)}}{2\times 5}$$

Whakareatia -20 ki te -9.

$$x=\frac{-12±\sqrt{144+180}}{2\times 5}$$

Tāpiri 144 ki te 180.

$$x=\frac{-12±\sqrt{324}}{2\times 5}$$

Tuhia te pūtakerua o te 324.

$$x=\frac{-12±18}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{-12±18}{10}$$

Nā, me whakaoti te whārite x=\frac{-12±18}{10} ina he tāpiri te ±. Tāpiri -12 ki te 18.

$$x=\frac{6}{10}$$

Whakahekea te hautanga \frac{6}{10} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{3}{5}$$

Nā, me whakaoti te whārite x=\frac{-12±18}{10} ina he tango te ±. Tango 18 mai i -12.

$$x=-\frac{30}{10}$$

Whakawehe -30 ki te 10.

$$x=-3$$

Kua oti te whārite te whakatau.

$$x=\frac{3}{5} x=-3$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te 4x-3 ki te x^{2}+x+3 ka whakakotahi i ngā kupu rite.

$$4x^{3}+x^{2}+9x-9=\left(2x+1\right)\left(2x^{2}-3x\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 2x+1 ki te 2x^{2}-3x ka whakakotahi i ngā kupu rite.

$$4x^{3}+x^{2}+9x-9=4x^{3}-4x^{2}-3x$$

Tangohia te 4x^{3} mai i ngā taha e rua.

$$4x^{3}+x^{2}+9x-9-4x^{3}=-4x^{2}-3x$$

Pahekotia te 4x^{3} me -4x^{3}, ka 0.

$$x^{2}+9x-9=-4x^{2}-3x$$

Me tāpiri te 4x^{2} ki ngā taha e rua.

$$x^{2}+9x-9+4x^{2}=-3x$$

Pahekotia te x^{2} me 4x^{2}, ka 5x^{2}.

$$5x^{2}+9x-9=-3x$$

Me tāpiri te 3x ki ngā taha e rua.

$$5x^{2}+9x-9+3x=0$$

Pahekotia te 9x me 3x, ka 12x.

$$5x^{2}+12x-9=0$$

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

$$5x^{2}+12x=9$$

Whakawehea ngā taha e rua ki te 5.

$$\frac{5x^{2}+12x}{5}=\frac{9}{5}$$

Mā te whakawehe ki te 5 ka wetekia te whakareanga ki te 5.

$$x^{2}+\frac{12}{5}x=\frac{9}{5}$$

Whakawehea te \frac{12}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{6}{5}. Nā, tāpiria te pūrua o te \frac{6}{5} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{12}{5}x+\left(\frac{6}{5}\right)^{2}=\frac{9}{5}+\left(\frac{6}{5}\right)^{2}$$

Pūruatia \frac{6}{5} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{12}{5}x+\frac{36}{25}=\frac{9}{5}+\frac{36}{25}$$

Tāpiri \frac{9}{5} ki te \frac{36}{25} 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.

$$x^{2}+\frac{12}{5}x+\frac{36}{25}=\frac{81}{25}$$

Tauwehea x^{2}+\frac{12}{5}x+\frac{36}{25}. 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}.

$$\left(x+\frac{6}{5}\right)^{2}=\frac{81}{25}$$

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

$$\sqrt{\left(x+\frac{6}{5}\right)^{2}}=\sqrt{\frac{81}{25}}$$

Whakarūnātia.

$$x+\frac{6}{5}=\frac{9}{5} x+\frac{6}{5}=-\frac{9}{5}$$

Me tango \frac{6}{5} mai i ngā taha e rua o te whārite.

$$x=\frac{3}{5} x=-3$$