Whakaoti i te 5x^2-5x-30=-20

$$5 x ^ { 2 } - 5 x - 30 = - 20$$

Whakaoti mō x

$$x=-1$$

$$x=2$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Me tāpiri te 20 ki ngā taha e rua.

$$5x^{2}-5x-30+20=0$$

Tāpirihia te -30 ki te 20, ka -10.

$$5x^{2}-5x-10=0$$

Whakawehea ngā taha e rua ki te 5.

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

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx-2. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-1 ab=1\left(-2\right)=-2$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-2 b=1$$

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

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

Whakatauwehea atu x i te x^{2}-2x.

$$x\left(x-2\right)+x-2$$

Whakatauwehea atu te kīanga pātahi x-2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-2\right)\left(x+1\right)$$

Hei kimi otinga whārite, me whakaoti te x-2=0 me te x+1=0.

$$x=2 x=-1$$

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.

$$5x^{2}-5x-30=-20$$

Me tāpiri 20 ki ngā taha e rua o te whārite.

$$5x^{2}-5x-30-\left(-20\right)=-20-\left(-20\right)$$

Mā te tango i te -20 i a ia ake anō ka toe ko te 0.

$$5x^{2}-5x-30-\left(-20\right)=0$$

Tango -20 mai i -30.

$$5x^{2}-5x-10=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 5 mō a, -5 mō b, me -10 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 5\left(-10\right)}}{2\times 5}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 5\left(-10\right)}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-\left(-5\right)±\sqrt{25-20\left(-10\right)}}{2\times 5}$$

Whakareatia -20 ki te -10.

$$x=\frac{-\left(-5\right)±\sqrt{25+200}}{2\times 5}$$

Tāpiri 25 ki te 200.

$$x=\frac{-\left(-5\right)±\sqrt{225}}{2\times 5}$$

Tuhia te pūtakerua o te 225.

$$x=\frac{-\left(-5\right)±15}{2\times 5}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±15}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{5±15}{10}$$

Nā, me whakaoti te whārite x=\frac{5±15}{10} ina he tāpiri te ±. Tāpiri 5 ki te 15.

$$x=\frac{20}{10}$$

Whakawehe 20 ki te 10.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{5±15}{10} ina he tango te ±. Tango 15 mai i 5.

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

Whakawehe -10 ki te 10.

$$x=-1$$

Kua oti te whārite te whakatau.

$$x=2 x=-1$$

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.

$$5x^{2}-5x-30=-20$$

Me tāpiri 30 ki ngā taha e rua o te whārite.

$$5x^{2}-5x-30-\left(-30\right)=-20-\left(-30\right)$$

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

$$5x^{2}-5x=-20-\left(-30\right)$$

Tango -30 mai i -20.

$$5x^{2}-5x=10$$

Whakawehea ngā taha e rua ki te 5.

$$\frac{5x^{2}-5x}{5}=\frac{10}{5}$$

Mā te whakawehe ki te 5 ka wetekia te whakareanga ki te 5.

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

Whakawehe -5 ki te 5.

$$x^{2}-x=\frac{10}{5}$$

Whakawehe 10 ki te 5.

$$x^{2}-x=2$$

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.

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

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

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

Tāpiri 2 ki te \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{9}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{9}{4}$$

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

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Whakarūnātia.

$$x-\frac{1}{2}=\frac{3}{2} x-\frac{1}{2}=-\frac{3}{2}$$

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

$$x=2 x=-1$$