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

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

Whakaoti mō x

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

$$x=2$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Me whakarea ngā taha e rua o te whārite ki te 4, arā, te tauraro pātahi he tino iti rawa te kitea o 2,4.

$$2\left(x+1\right)\left(x-3\right)+4x=x$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te x+1.

$$\left(2x+2\right)\left(x-3\right)+4x=x$$

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

$$2x^{2}-4x-6+4x=x$$

Pahekotia te -4x me 4x, ka 0.

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

Tangohia te x mai i ngā taha e rua.

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

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

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

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 2x^{2}+ax+bx-6. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-1 ab=2\left(-6\right)=-12$$

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 -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 -1.

$$a=-4 b=3$$

Tuhia anō te 2x^{2}-x-6 hei \left(2x^{2}-4x\right)+\left(3x-6\right).

$$\left(2x^{2}-4x\right)+\left(3x-6\right)$$

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

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

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

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

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

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Me whakarea ngā taha e rua o te whārite ki te 4, arā, te tauraro pātahi he tino iti rawa te kitea o 2,4.

$$2\left(x+1\right)\left(x-3\right)+4x=x$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te x+1.

$$\left(2x+2\right)\left(x-3\right)+4x=x$$

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

$$2x^{2}-4x-6+4x=x$$

Pahekotia te -4x me 4x, ka 0.

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

Tangohia te x mai i ngā taha e rua.

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

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.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, -1 mō b, me -6 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 2\left(-6\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-1\right)±\sqrt{1-8\left(-6\right)}}{2\times 2}$$

Whakareatia -8 ki te -6.

$$x=\frac{-\left(-1\right)±\sqrt{1+48}}{2\times 2}$$

Tāpiri 1 ki te 48.

$$x=\frac{-\left(-1\right)±\sqrt{49}}{2\times 2}$$

Tuhia te pūtakerua o te 49.

$$x=\frac{-\left(-1\right)±7}{2\times 2}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±7}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{1±7}{4}$$

Nā, me whakaoti te whārite x=\frac{1±7}{4} ina he tāpiri te ±. Tāpiri 1 ki te 7.

$$x=\frac{8}{4}$$

Whakawehe 8 ki te 4.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{1±7}{4} ina he tango te ±. Tango 7 mai i 1.

$$x=-\frac{6}{4}$$

Whakahekea te hautanga \frac{-6}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Me whakarea ngā taha e rua o te whārite ki te 4, arā, te tauraro pātahi he tino iti rawa te kitea o 2,4.

$$2\left(x+1\right)\left(x-3\right)+4x=x$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te x+1.

$$\left(2x+2\right)\left(x-3\right)+4x=x$$

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

$$2x^{2}-4x-6+4x=x$$

Pahekotia te -4x me 4x, ka 0.

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

Tangohia te x mai i ngā taha e rua.

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

Me tāpiri te 6 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

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

Whakawehea ngā taha e rua ki te 2.

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

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

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

Whakawehe 6 ki te 2.

$$x^{2}-\frac{1}{2}x=3$$

Whakawehea te -\frac{1}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{4}. Nā, tāpiria te pūrua o te -\frac{1}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia -\frac{1}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=3+\frac{1}{16}$$

Tāpiri 3 ki te \frac{1}{16}.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{49}{16}$$

Tauwehea x^{2}-\frac{1}{2}x+\frac{1}{16}. 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}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{49}{16}$$

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

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Whakarūnātia.

$$x-\frac{1}{4}=\frac{7}{4} x-\frac{1}{4}=-\frac{7}{4}$$

Me tāpiri \frac{1}{4} ki ngā taha e rua o te whārite.

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