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

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

Whakaoti mō x

$$x=-3$$

$$x=2$$

Ngā Upane Whakamahi Tauwehe

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+2\right)^{2}.

$$x^{2}+4x+4-3\left(x+2\right)-4=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te -3 ki te x+2.

$$x^{2}+4x+4-3x-6-4=0$$

Pahekotia te 4x me -3x, ka x.

$$x^{2}+x+4-6-4=0$$

Tangohia te 6 i te 4, ka -2.

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

Tangohia te 4 i te -2, ka -6.

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

Hei whakaoti i te whārite, whakatauwehea te x^{2}+x-6 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.

$$a+b=1 ab=-6$$

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 -6.

$$-1,6 -2,3$$

Tātaihia te tapeke mō ia takirua.

$$-1+6=5 -2+3=1$$

Ko te otinga te takirua ka hoatu i te tapeke 1.

$$a=-2 b=3$$

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

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

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

$$x=2 x=-3$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+2\right)^{2}.

$$x^{2}+4x+4-3\left(x+2\right)-4=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te -3 ki te x+2.

$$x^{2}+4x+4-3x-6-4=0$$

Pahekotia te 4x me -3x, ka x.

$$x^{2}+x+4-6-4=0$$

Tangohia te 6 i te 4, ka -2.

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

Tangohia te 4 i te -2, ka -6.

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

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

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

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 -6.

$$-1,6 -2,3$$

Tātaihia te tapeke mō ia takirua.

$$-1+6=5 -2+3=1$$

Ko te otinga te takirua ka hoatu i te tapeke 1.

$$a=-2 b=3$$

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

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

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

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

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

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

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

$$x=2 x=-3$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+2\right)^{2}.

$$x^{2}+4x+4-3\left(x+2\right)-4=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te -3 ki te x+2.

$$x^{2}+4x+4-3x-6-4=0$$

Pahekotia te 4x me -3x, ka x.

$$x^{2}+x+4-6-4=0$$

Tangohia te 6 i te 4, ka -2.

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

Tangohia te 4 i te -2, ka -6.

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

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

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

Pūrua 1.

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

Whakareatia -4 ki te -6.

$$x=\frac{-1±\sqrt{1+24}}{2}$$

Tāpiri 1 ki te 24.

$$x=\frac{-1±\sqrt{25}}{2}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-1±5}{2}$$

Nā, me whakaoti te whārite x=\frac{-1±5}{2} ina he tāpiri te ±. Tāpiri -1 ki te 5.

$$x=\frac{4}{2}$$

Whakawehe 4 ki te 2.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{-1±5}{2} ina he tango te ±. Tango 5 mai i -1.

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

Whakawehe -6 ki te 2.

$$x=-3$$

Kua oti te whārite te whakatau.

$$x=2 x=-3$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+2\right)^{2}.

$$x^{2}+4x+4-3\left(x+2\right)-4=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te -3 ki te x+2.

$$x^{2}+4x+4-3x-6-4=0$$

Pahekotia te 4x me -3x, ka x.

$$x^{2}+x+4-6-4=0$$

Tangohia te 6 i te 4, ka -2.

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

Tangohia te 4 i te -2, ka -6.

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

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

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

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

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=6+\left(\frac{1}{2}\right)^{2}$$

Pūruatia \frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+x+\frac{1}{4}=6+\frac{1}{4}$$

Tāpiri 6 ki te \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{25}{4}$$

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

$$\left(x+\frac{1}{2}\right)^{2}=\frac{25}{4}$$

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

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Whakarūnātia.

$$x+\frac{1}{2}=\frac{5}{2} x+\frac{1}{2}=-\frac{5}{2}$$

Me tango \frac{1}{2} mai i ngā taha e rua o te whārite.

$$x=2 x=-3$$