Whakaoti i te s^2+13s+42=0

$$s ^ { 2 } + 13 s + 42 = 0$$

Whakaoti mō s

$$s=-7$$

$$s=-6$$

Ngā Upane Whakamahi Tauwehe

Hei whakaoti i te whārite, whakatauwehea te s^{2}+13s+42 mā te whakamahi i te tātai s^{2}+\left(a+b\right)s+ab=\left(s+a\right)\left(s+b\right). Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=13 ab=42$$

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 42.

$$1,42 2,21 3,14 6,7$$

Tātaihia te tapeke mō ia takirua.

$$1+42=43 2+21=23 3+14=17 6+7=13$$

Ko te otinga te takirua ka hoatu i te tapeke 13.

$$a=6 b=7$$

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

$$\left(s+6\right)\left(s+7\right)$$

Hei kimi otinga whārite, me whakaoti te s+6=0 me te s+7=0.

$$s=-6 s=-7$$

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 s^{2}+as+bs+42. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=13 ab=1\times 42=42$$

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 42.

$$1,42 2,21 3,14 6,7$$

Tātaihia te tapeke mō ia takirua.

$$1+42=43 2+21=23 3+14=17 6+7=13$$

Ko te otinga te takirua ka hoatu i te tapeke 13.

$$a=6 b=7$$

Tuhia anō te s^{2}+13s+42 hei \left(s^{2}+6s\right)+\left(7s+42\right).

$$\left(s^{2}+6s\right)+\left(7s+42\right)$$

Tauwehea te s i te tuatahi me te 7 i te rōpū tuarua.

$$s\left(s+6\right)+7\left(s+6\right)$$

Whakatauwehea atu te kīanga pātahi s+6 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(s+6\right)\left(s+7\right)$$

Hei kimi otinga whārite, me whakaoti te s+6=0 me te s+7=0.

$$s=-6 s=-7$$

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.

$$s^{2}+13s+42=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 13 mō b, me 42 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$s=\frac{-13±\sqrt{13^{2}-4\times 42}}{2}$$

Pūrua 13.

$$s=\frac{-13±\sqrt{169-4\times 42}}{2}$$

Whakareatia -4 ki te 42.

$$s=\frac{-13±\sqrt{169-168}}{2}$$

Tāpiri 169 ki te -168.

$$s=\frac{-13±\sqrt{1}}{2}$$

Tuhia te pūtakerua o te 1.

$$s=\frac{-13±1}{2}$$

Nā, me whakaoti te whārite s=\frac{-13±1}{2} ina he tāpiri te ±. Tāpiri -13 ki te 1.

$$s=-\frac{12}{2}$$

Whakawehe -12 ki te 2.

$$s=-6$$

Nā, me whakaoti te whārite s=\frac{-13±1}{2} ina he tango te ±. Tango 1 mai i -13.

$$s=-\frac{14}{2}$$

Whakawehe -14 ki te 2.

$$s=-7$$

Kua oti te whārite te whakatau.

$$s=-6 s=-7$$

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.

$$s^{2}+13s+42=0$$

Me tango 42 mai i ngā taha e rua o te whārite.

$$s^{2}+13s+42-42=-42$$

Mā te tango i te 42 i a ia ake anō ka toe ko te 0.

$$s^{2}+13s=-42$$

Whakawehea te 13, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{13}{2}. Nā, tāpiria te pūrua o te \frac{13}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$s^{2}+13s+\left(\frac{13}{2}\right)^{2}=-42+\left(\frac{13}{2}\right)^{2}$$

Pūruatia \frac{13}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$s^{2}+13s+\frac{169}{4}=-42+\frac{169}{4}$$

Tāpiri -42 ki te \frac{169}{4}.

$$s^{2}+13s+\frac{169}{4}=\frac{1}{4}$$

Tauwehea s^{2}+13s+\frac{169}{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}.

$$\left(s+\frac{13}{2}\right)^{2}=\frac{1}{4}$$

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

$$\sqrt{\left(s+\frac{13}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Whakarūnātia.

$$s+\frac{13}{2}=\frac{1}{2} s+\frac{13}{2}=-\frac{1}{2}$$

Me tango \frac{13}{2} mai i ngā taha e rua o te whārite.

$$s=-6 s=-7$$