Whakaoti i te 3x+5=x^2+1

$$3 x + 5 = x ^ { 2 } + 1$$

Whakaoti mō x

$$x=-1$$

$$x=4$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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

$$3x+5-x^{2}=1$$

Tangohia te 1 mai i ngā taha e rua.

$$3x+5-x^{2}-1=0$$

Tangohia te 1 i te 5, ka 4.

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

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

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

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

$$a+b=3 ab=-4=-4$$

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 -4.

$$-1,4 -2,2$$

Tātaihia te tapeke mō ia takirua.

$$-1+4=3 -2+2=0$$

Ko te otinga te takirua ka hoatu i te tapeke 3.

$$a=4 b=-1$$

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

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

Tauwehea te -x i te tuatahi me te -1 i te rōpū tuarua.

$$-x\left(x-4\right)-\left(x-4\right)$$

Whakatauwehea atu te kīanga pātahi x-4 mā te whakamahi i te āhuatanga tātai tohatoha.

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

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

$$x=4 x=-1$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

$$3x+5-x^{2}=1$$

Tangohia te 1 mai i ngā taha e rua.

$$3x+5-x^{2}-1=0$$

Tangohia te 1 i te 5, ka 4.

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

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.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -1 mō a, 3 mō b, me 4 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua 3.

$$x=\frac{-3±\sqrt{9-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Whakareatia -4 ki te -1.

$$x=\frac{-3±\sqrt{9+4\times 4}}{2\left(-1\right)}$$

Whakareatia 4 ki te 4.

$$x=\frac{-3±\sqrt{9+16}}{2\left(-1\right)}$$

Tāpiri 9 ki te 16.

$$x=\frac{-3±\sqrt{25}}{2\left(-1\right)}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-3±5}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{-3±5}{-2}$$

Nā, me whakaoti te whārite x=\frac{-3±5}{-2} ina he tāpiri te ±. Tāpiri -3 ki te 5.

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

Whakawehe 2 ki te -2.

$$x=-1$$

Nā, me whakaoti te whārite x=\frac{-3±5}{-2} ina he tango te ±. Tango 5 mai i -3.

$$x=-\frac{8}{-2}$$

Whakawehe -8 ki te -2.

$$x=4$$

Kua oti te whārite te whakatau.

$$x=-1 x=4$$

Ngā Upane mō te Whakaoti i te Pūrua

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

$$3x+5-x^{2}=1$$

Tangohia te 5 mai i ngā taha e rua.

$$3x-x^{2}=1-5$$

Tangohia te 5 i te 1, ka -4.

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

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.

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

Whakawehea ngā taha e rua ki te -1.

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

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

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

Whakawehe 3 ki te -1.

$$x^{2}-3x=-\frac{4}{-1}$$

Whakawehe -4 ki te -1.

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

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.

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

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

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

Tāpiri 4 ki te \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{25}{4}$$

Tauwehea x^{2}-3x+\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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{25}{4}$$

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

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Whakarūnātia.

$$x-\frac{3}{2}=\frac{5}{2} x-\frac{3}{2}=-\frac{5}{2}$$

Me tāpiri \frac{3}{2} ki ngā taha e rua o te whārite.

$$x=4 x=-1$$