Whakaoti i te x^2+x-650=0

$${ x }^{ 2 } +x-650=0$$

Whakaoti mō x

$$x=-26$$

$$x=25$$

Ngā Upane Whakamahi Tauwehe

Hei whakaoti i te whārite, whakatauwehea te x^{2}+x-650 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.

$$a+b=1 ab=-650$$

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 -650.

$$-1,650 -2,325 -5,130 -10,65 -13,50 -25,26$$

Tātaihia te tapeke mō ia takirua.

$$-1+650=649 -2+325=323 -5+130=125 -10+65=55 -13+50=37 -25+26=1$$

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

$$a=-25 b=26$$

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

$$\left(x-25\right)\left(x+26\right)$$

Hei kimi otinga whārite, me whakaoti te x-25=0 me te x+26=0.

$$x=25 x=-26$$

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-650. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=1 ab=1\left(-650\right)=-650$$

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 -650.

$$-1,650 -2,325 -5,130 -10,65 -13,50 -25,26$$

Tātaihia te tapeke mō ia takirua.

$$-1+650=649 -2+325=323 -5+130=125 -10+65=55 -13+50=37 -25+26=1$$

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

$$a=-25 b=26$$

Tuhia anō te x^{2}+x-650 hei \left(x^{2}-25x\right)+\left(26x-650\right).

$$\left(x^{2}-25x\right)+\left(26x-650\right)$$

Tauwehea te x i te tuatahi me te 26 i te rōpū tuarua.

$$x\left(x-25\right)+26\left(x-25\right)$$

Whakatauwehea atu te kīanga pātahi x-25 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-25\right)\left(x+26\right)$$

Hei kimi otinga whārite, me whakaoti te x-25=0 me te x+26=0.

$$x=25 x=-26$$

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.

$$x^{2}+x-650=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 1 mō b, me -650 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-650\right)}}{2}$$

Pūrua 1.

$$x=\frac{-1±\sqrt{1-4\left(-650\right)}}{2}$$

Whakareatia -4 ki te -650.

$$x=\frac{-1±\sqrt{1+2600}}{2}$$

Tāpiri 1 ki te 2600.

$$x=\frac{-1±\sqrt{2601}}{2}$$

Tuhia te pūtakerua o te 2601.

$$x=\frac{-1±51}{2}$$

Nā, me whakaoti te whārite x=\frac{-1±51}{2} ina he tāpiri te ±. Tāpiri -1 ki te 51.

$$x=\frac{50}{2}$$

Whakawehe 50 ki te 2.

$$x=25$$

Nā, me whakaoti te whārite x=\frac{-1±51}{2} ina he tango te ±. Tango 51 mai i -1.

$$x=-\frac{52}{2}$$

Whakawehe -52 ki te 2.

$$x=-26$$

Kua oti te whārite te whakatau.

$$x=25 x=-26$$

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.

$$x^{2}+x-650=0$$

Me tāpiri 650 ki ngā taha e rua o te whārite.

$$x^{2}+x-650-\left(-650\right)=-\left(-650\right)$$

Mā te tango i te -650 i a ia ake anō ka toe ko te 0.

$$x^{2}+x=-\left(-650\right)$$

Tango -650 mai i 0.

$$x^{2}+x=650$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=650+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=650+\frac{1}{4}$$

Tāpiri 650 ki te \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{2601}{4}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{2601}{4}$$

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

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{2601}{4}}$$

Whakarūnātia.

$$x+\frac{1}{2}=\frac{51}{2} x+\frac{1}{2}=-\frac{51}{2}$$

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

$$x=25 x=-26$$