Whakaoti i te b^2-11b+30=0

$$b ^ { 2 } - 11 b + 30 = 0$$

Whakaoti mō b

$$b=5$$

$$b=6$$

Ngā Upane Whakamahi Tauwehe

Hei whakaoti i te whārite, whakatauwehea te b^{2}-11b+30 mā te whakamahi i te tātai b^{2}+\left(a+b\right)b+ab=\left(b+a\right)\left(b+b\right). Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-11 ab=30$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Tātaihia te tapeke mō ia takirua.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

Ko te otinga te takirua ka hoatu i te tapeke -11.

$$a=-6 b=-5$$

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

$$\left(b-6\right)\left(b-5\right)$$

Hei kimi otinga whārite, me whakaoti te b-6=0 me te b-5=0.

$$b=6 b=5$$

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 b^{2}+ab+bb+30. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-11 ab=1\times 30=30$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Tātaihia te tapeke mō ia takirua.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

Ko te otinga te takirua ka hoatu i te tapeke -11.

$$a=-6 b=-5$$

Tuhia anō te b^{2}-11b+30 hei \left(b^{2}-6b\right)+\left(-5b+30\right).

$$\left(b^{2}-6b\right)+\left(-5b+30\right)$$

Tauwehea te b i te tuatahi me te -5 i te rōpū tuarua.

$$b\left(b-6\right)-5\left(b-6\right)$$

Whakatauwehea atu te kīanga pātahi b-6 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(b-6\right)\left(b-5\right)$$

Hei kimi otinga whārite, me whakaoti te b-6=0 me te b-5=0.

$$b=6 b=5$$

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.

$$b^{2}-11b+30=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -11 mō b, me 30 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 30}}{2}$$

Pūrua -11.

$$b=\frac{-\left(-11\right)±\sqrt{121-4\times 30}}{2}$$

Whakareatia -4 ki te 30.

$$b=\frac{-\left(-11\right)±\sqrt{121-120}}{2}$$

Tāpiri 121 ki te -120.

$$b=\frac{-\left(-11\right)±\sqrt{1}}{2}$$

Tuhia te pūtakerua o te 1.

$$b=\frac{-\left(-11\right)±1}{2}$$

Ko te tauaro o -11 ko 11.

$$b=\frac{11±1}{2}$$

Nā, me whakaoti te whārite b=\frac{11±1}{2} ina he tāpiri te ±. Tāpiri 11 ki te 1.

$$b=\frac{12}{2}$$

Whakawehe 12 ki te 2.

$$b=6$$

Nā, me whakaoti te whārite b=\frac{11±1}{2} ina he tango te ±. Tango 1 mai i 11.

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

Whakawehe 10 ki te 2.

$$b=5$$

Kua oti te whārite te whakatau.

$$b=6 b=5$$

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.

$$b^{2}-11b+30=0$$

Me tango 30 mai i ngā taha e rua o te whārite.

$$b^{2}-11b+30-30=-30$$

Mā te tango i te 30 i a ia ake anō ka toe ko te 0.

$$b^{2}-11b=-30$$

Whakawehea te -11, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{11}{2}. Nā, tāpiria te pūrua o te -\frac{11}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$b^{2}-11b+\left(-\frac{11}{2}\right)^{2}=-30+\left(-\frac{11}{2}\right)^{2}$$

Pūruatia -\frac{11}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$b^{2}-11b+\frac{121}{4}=-30+\frac{121}{4}$$

Tāpiri -30 ki te \frac{121}{4}.

$$b^{2}-11b+\frac{121}{4}=\frac{1}{4}$$

Tauwehea b^{2}-11b+\frac{121}{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}.

$$\left(b-\frac{11}{2}\right)^{2}=\frac{1}{4}$$

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

$$\sqrt{\left(b-\frac{11}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Whakarūnātia.

$$b-\frac{11}{2}=\frac{1}{2} b-\frac{11}{2}=-\frac{1}{2}$$

Me tāpiri \frac{11}{2} ki ngā taha e rua o te whārite.

$$b=6 b=5$$