Whakaoti i te 5x^2-40x-45=0

$$5 x ^ { 2 } - 40 x - 45 = 0$$

Whakaoti mō x

$$x=-1$$

$$x=9$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakawehea ngā taha e rua ki te 5.

$$x^{2}-8x-9=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx-9. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-8 ab=1\left(-9\right)=-9$$

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. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -9.

$$1,-9 3,-3$$

Tātaihia te tapeke mō ia takirua.

$$1-9=-8 3-3=0$$

Ko te otinga te takirua ka hoatu i te tapeke -8.

$$a=-9 b=1$$

Tuhia anō te x^{2}-8x-9 hei \left(x^{2}-9x\right)+\left(x-9\right).

$$\left(x^{2}-9x\right)+\left(x-9\right)$$

Whakatauwehea atu x i te x^{2}-9x.

$$x\left(x-9\right)+x-9$$

Whakatauwehea atu te kīanga pātahi x-9 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-9\right)\left(x+1\right)$$

Hei kimi otinga whārite, me whakaoti te x-9=0 me te x+1=0.

$$x=9 x=-1$$

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.

$$5x^{2}-40x-45=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 5 mō a, -40 mō b, me -45 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 5\left(-45\right)}}{2\times 5}$$

Pūrua -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 5\left(-45\right)}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-\left(-40\right)±\sqrt{1600-20\left(-45\right)}}{2\times 5}$$

Whakareatia -20 ki te -45.

$$x=\frac{-\left(-40\right)±\sqrt{1600+900}}{2\times 5}$$

Tāpiri 1600 ki te 900.

$$x=\frac{-\left(-40\right)±\sqrt{2500}}{2\times 5}$$

Tuhia te pūtakerua o te 2500.

$$x=\frac{-\left(-40\right)±50}{2\times 5}$$

Ko te tauaro o -40 ko 40.

$$x=\frac{40±50}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{40±50}{10}$$

Nā, me whakaoti te whārite x=\frac{40±50}{10} ina he tāpiri te ±. Tāpiri 40 ki te 50.

$$x=\frac{90}{10}$$

Whakawehe 90 ki te 10.

$$x=9$$

Nā, me whakaoti te whārite x=\frac{40±50}{10} ina he tango te ±. Tango 50 mai i 40.

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

Whakawehe -10 ki te 10.

$$x=-1$$

Kua oti te whārite te whakatau.

$$x=9 x=-1$$

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.

$$5x^{2}-40x-45=0$$

Me tāpiri 45 ki ngā taha e rua o te whārite.

$$5x^{2}-40x-45-\left(-45\right)=-\left(-45\right)$$

Mā te tango i te -45 i a ia ake anō ka toe ko te 0.

$$5x^{2}-40x=-\left(-45\right)$$

Tango -45 mai i 0.

$$5x^{2}-40x=45$$

Whakawehea ngā taha e rua ki te 5.

$$\frac{5x^{2}-40x}{5}=\frac{45}{5}$$

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

$$x^{2}+\left(-\frac{40}{5}\right)x=\frac{45}{5}$$

Whakawehe -40 ki te 5.

$$x^{2}-8x=\frac{45}{5}$$

Whakawehe 45 ki te 5.

$$x^{2}-8x=9$$

Whakawehea te -8, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -4. Nā, tāpiria te pūrua o te -4 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-8x+\left(-4\right)^{2}=9+\left(-4\right)^{2}$$

Pūrua -4.

$$x^{2}-8x+16=9+16$$

Tāpiri 9 ki te 16.

$$x^{2}-8x+16=25$$

Tauwehea x^{2}-8x+16. 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}.

$$\left(x-4\right)^{2}=25$$

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

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{25}$$

Whakarūnātia.

$$x-4=5 x-4=-5$$

Me tāpiri 4 ki ngā taha e rua o te whārite.

$$x=9 x=-1$$