Whakaoti i te x^2+5x-50=0

$$x ^ { 2 } + 5 x - 50 = 0$$

Whakaoti mō x

$$x=-10$$

$$x=5$$

Ngā Upane Whakamahi Tauwehe

Hei whakaoti i te whārite, whakatauwehea te x^{2}+5x-50 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.

$$a+b=5 ab=-50$$

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 -50.

$$-1,50 -2,25 -5,10$$

Tātaihia te tapeke mō ia takirua.

$$-1+50=49 -2+25=23 -5+10=5$$

Ko te otinga te takirua ka hoatu i te tapeke 5.

$$a=-5 b=10$$

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

$$\left(x-5\right)\left(x+10\right)$$

Hei kimi otinga whārite, me whakaoti te x-5=0 me te x+10=0.

$$x=5 x=-10$$

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-50. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=5 ab=1\left(-50\right)=-50$$

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 -50.

$$-1,50 -2,25 -5,10$$

Tātaihia te tapeke mō ia takirua.

$$-1+50=49 -2+25=23 -5+10=5$$

Ko te otinga te takirua ka hoatu i te tapeke 5.

$$a=-5 b=10$$

Tuhia anō te x^{2}+5x-50 hei \left(x^{2}-5x\right)+\left(10x-50\right).

$$\left(x^{2}-5x\right)+\left(10x-50\right)$$

Tauwehea te x i te tuatahi me te 10 i te rōpū tuarua.

$$x\left(x-5\right)+10\left(x-5\right)$$

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

$$\left(x-5\right)\left(x+10\right)$$

Hei kimi otinga whārite, me whakaoti te x-5=0 me te x+10=0.

$$x=5 x=-10$$

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.

$$x^{2}+5x-50=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 5 mō b, me -50 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-50\right)}}{2}$$

Pūrua 5.

$$x=\frac{-5±\sqrt{25-4\left(-50\right)}}{2}$$

Whakareatia -4 ki te -50.

$$x=\frac{-5±\sqrt{25+200}}{2}$$

Tāpiri 25 ki te 200.

$$x=\frac{-5±\sqrt{225}}{2}$$

Tuhia te pūtakerua o te 225.

$$x=\frac{-5±15}{2}$$

Nā, me whakaoti te whārite x=\frac{-5±15}{2} ina he tāpiri te ±. Tāpiri -5 ki te 15.

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

Whakawehe 10 ki te 2.

$$x=5$$

Nā, me whakaoti te whārite x=\frac{-5±15}{2} ina he tango te ±. Tango 15 mai i -5.

$$x=-\frac{20}{2}$$

Whakawehe -20 ki te 2.

$$x=-10$$

Kua oti te whārite te whakatau.

$$x=5 x=-10$$

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.

$$x^{2}+5x-50=0$$

Me tāpiri 50 ki ngā taha e rua o te whārite.

$$x^{2}+5x-50-\left(-50\right)=-\left(-50\right)$$

Mā te tango i te -50 i a ia ake anō ka toe ko te 0.

$$x^{2}+5x=-\left(-50\right)$$

Tango -50 mai i 0.

$$x^{2}+5x=50$$

Whakawehea te 5, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{5}{2}. Nā, tāpiria te pūrua o te \frac{5}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=50+\left(\frac{5}{2}\right)^{2}$$

Pūruatia \frac{5}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+5x+\frac{25}{4}=50+\frac{25}{4}$$

Tāpiri 50 ki te \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{225}{4}$$

Tauwehea x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{225}{4}$$

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

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{225}{4}}$$

Whakarūnātia.

$$x+\frac{5}{2}=\frac{15}{2} x+\frac{5}{2}=-\frac{15}{2}$$

Me tango \frac{5}{2} mai i ngā taha e rua o te whārite.

$$x=5 x=-10$$