Whakaoti i te x^2+25x+100=0

$${ x }^{ 2 } +25x+100=0$$

Whakaoti mō x

$$x=-20$$

$$x=-5$$

Ngā Upane Whakamahi Tauwehe

Hei whakaoti i te whārite, whakatauwehea te x^{2}+25x+100 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.

$$a+b=25 ab=100$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 100.

$$1,100 2,50 4,25 5,20 10,10$$

Tātaihia te tapeke mō ia takirua.

$$1+100=101 2+50=52 4+25=29 5+20=25 10+10=20$$

Ko te otinga te takirua ka hoatu i te tapeke 25.

$$a=5 b=20$$

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

$$\left(x+5\right)\left(x+20\right)$$

Hei kimi otinga whārite, me whakaoti te x+5=0 me te x+20=0.

$$x=-5 x=-20$$

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+100. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=25 ab=1\times 100=100$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 100.

$$1,100 2,50 4,25 5,20 10,10$$

Tātaihia te tapeke mō ia takirua.

$$1+100=101 2+50=52 4+25=29 5+20=25 10+10=20$$

Ko te otinga te takirua ka hoatu i te tapeke 25.

$$a=5 b=20$$

Tuhia anō te x^{2}+25x+100 hei \left(x^{2}+5x\right)+\left(20x+100\right).

$$\left(x^{2}+5x\right)+\left(20x+100\right)$$

Tauwehea te x i te tuatahi me te 20 i te rōpū tuarua.

$$x\left(x+5\right)+20\left(x+5\right)$$

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

$$\left(x+5\right)\left(x+20\right)$$

Hei kimi otinga whārite, me whakaoti te x+5=0 me te x+20=0.

$$x=-5 x=-20$$

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.

$$x^{2}+25x+100=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 25 mō b, me 100 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-25±\sqrt{25^{2}-4\times 100}}{2}$$

Pūrua 25.

$$x=\frac{-25±\sqrt{625-4\times 100}}{2}$$

Whakareatia -4 ki te 100.

$$x=\frac{-25±\sqrt{625-400}}{2}$$

Tāpiri 625 ki te -400.

$$x=\frac{-25±\sqrt{225}}{2}$$

Tuhia te pūtakerua o te 225.

$$x=\frac{-25±15}{2}$$

Nā, me whakaoti te whārite x=\frac{-25±15}{2} ina he tāpiri te ±. Tāpiri -25 ki te 15.

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

Whakawehe -10 ki te 2.

$$x=-5$$

Nā, me whakaoti te whārite x=\frac{-25±15}{2} ina he tango te ±. Tango 15 mai i -25.

$$x=-\frac{40}{2}$$

Whakawehe -40 ki te 2.

$$x=-20$$

Kua oti te whārite te whakatau.

$$x=-5 x=-20$$

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.

$$x^{2}+25x+100=0$$

Me tango 100 mai i ngā taha e rua o te whārite.

$$x^{2}+25x+100-100=-100$$

Mā te tango i te 100 i a ia ake anō ka toe ko te 0.

$$x^{2}+25x=-100$$

Whakawehea te 25, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{25}{2}. Nā, tāpiria te pūrua o te \frac{25}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+25x+\left(\frac{25}{2}\right)^{2}=-100+\left(\frac{25}{2}\right)^{2}$$

Pūruatia \frac{25}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+25x+\frac{625}{4}=-100+\frac{625}{4}$$

Tāpiri -100 ki te \frac{625}{4}.

$$x^{2}+25x+\frac{625}{4}=\frac{225}{4}$$

Tauwehea x^{2}+25x+\frac{625}{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}.

$$\left(x+\frac{25}{2}\right)^{2}=\frac{225}{4}$$

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

$$\sqrt{\left(x+\frac{25}{2}\right)^{2}}=\sqrt{\frac{225}{4}}$$

Whakarūnātia.

$$x+\frac{25}{2}=\frac{15}{2} x+\frac{25}{2}=-\frac{15}{2}$$

Me tango \frac{25}{2} mai i ngā taha e rua o te whārite.

$$x=-5 x=-20$$