Whakaoti i te (x+3)(x-4)-8=0

$$(x+3)(x-4)-8=0$$

Whakaoti mō x

$$x=-4$$

$$x=5$$

Ngā Upane Whakamahi Tauwehe

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

$$x^{2}-x-12-8=0$$

Tangohia te 8 i te -12, ka -20.

$$x^{2}-x-20=0$$

Hei whakaoti i te whārite, whakatauwehea te x^{2}-x-20 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.

$$a+b=-1 ab=-20$$

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 -20.

$$1,-20 2,-10 4,-5$$

Tātaihia te tapeke mō ia takirua.

$$1-20=-19 2-10=-8 4-5=-1$$

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

$$a=-5 b=4$$

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

$$\left(x-5\right)\left(x+4\right)$$

Hei kimi otinga whārite, me whakaoti te x-5=0 me te x+4=0.

$$x=5 x=-4$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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

$$x^{2}-x-12-8=0$$

Tangohia te 8 i te -12, ka -20.

$$x^{2}-x-20=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx-20. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-1 ab=1\left(-20\right)=-20$$

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 -20.

$$1,-20 2,-10 4,-5$$

Tātaihia te tapeke mō ia takirua.

$$1-20=-19 2-10=-8 4-5=-1$$

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

$$a=-5 b=4$$

Tuhia anō te x^{2}-x-20 hei \left(x^{2}-5x\right)+\left(4x-20\right).

$$\left(x^{2}-5x\right)+\left(4x-20\right)$$

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

$$x\left(x-5\right)+4\left(x-5\right)$$

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

$$\left(x-5\right)\left(x+4\right)$$

Hei kimi otinga whārite, me whakaoti te x-5=0 me te x+4=0.

$$x=5 x=-4$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

$$x^{2}-x-12-8=0$$

Tangohia te 8 i te -12, ka -20.

$$x^{2}-x-20=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -1 mō b, me -20 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-20\right)}}{2}$$

Whakareatia -4 ki te -20.

$$x=\frac{-\left(-1\right)±\sqrt{1+80}}{2}$$

Tāpiri 1 ki te 80.

$$x=\frac{-\left(-1\right)±\sqrt{81}}{2}$$

Tuhia te pūtakerua o te 81.

$$x=\frac{-\left(-1\right)±9}{2}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±9}{2}$$

Nā, me whakaoti te whārite x=\frac{1±9}{2} ina he tāpiri te ±. Tāpiri 1 ki te 9.

$$x=\frac{10}{2}$$

Whakawehe 10 ki te 2.

$$x=5$$

Nā, me whakaoti te whārite x=\frac{1±9}{2} ina he tango te ±. Tango 9 mai i 1.

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

Whakawehe -8 ki te 2.

$$x=-4$$

Kua oti te whārite te whakatau.

$$x=5 x=-4$$

Ngā Upane mō te Whakaoti i te Pūrua

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

$$x^{2}-x-12-8=0$$

Tangohia te 8 i te -12, ka -20.

$$x^{2}-x-20=0$$

Me tāpiri te 20 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$x^{2}-x=20$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=20+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=20+\frac{1}{4}$$

Tāpiri 20 ki te \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{81}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{81}{4}$$

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

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Whakarūnātia.

$$x-\frac{1}{2}=\frac{9}{2} x-\frac{1}{2}=-\frac{9}{2}$$

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

$$x=5 x=-4$$