Whakaoti i te x^2+3x-4=0

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

Whakaoti mō x

$$x=-4$$

$$x=1$$

Ngā Upane Whakamahi Tauwehe

Hei whakaoti i te whārite, whakatauwehea te x^{2}+3x-4 mā te whakamahi i te tātai x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

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

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.

$$a=-1 b=4$$

Me tuhi anō te kīanga whakatauwehe \left(x+a\right)\left(x+b\right) mā ngā uara i tātaihia.

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

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

$$x=1 x=-4$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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.

$$a+b=3 ab=1\left(-4\right)=-4$$

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.

$$a=-1 b=4$$

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

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

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

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

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

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

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

$$x=1 x=-4$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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}.

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

Pūrua 3.

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

Whakareatia -4 ki te -4.

$$x=\frac{-3±\sqrt{9+16}}{2}$$

Tāpiri 9 ki te 16.

$$x=\frac{-3±\sqrt{25}}{2}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-3±5}{2}$$

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

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.

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

Me tāpiri 4 ki ngā taha e rua o te whārite.

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

Mā te tango i te -4 i a ia ake anō ka toe ko te 0.

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

Tango -4 mai i 0.

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

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 te x^{2}+3x+\frac{9}{4}. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe 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 tango \frac{3}{2} mai i ngā taha e rua o te whārite.

$$x=1 x=-4$$