Whakaoti i te -80x^2+50x-5=0

$$-80 { x }^{ 2 } +50x-5=0$$

Whakaoti mō x

$$x=\frac{1}{8}=0.125$$

$$x=\frac{1}{2}=0.5$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakawehea ngā taha e rua ki te 5.

$$-16x^{2}+10x-1=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei -16x^{2}+ax+bx-1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=10 ab=-16\left(-1\right)=16$$

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 16.

$$1,16 2,8 4,4$$

Tātaihia te tapeke mō ia takirua.

$$1+16=17 2+8=10 4+4=8$$

Ko te otinga te takirua ka hoatu i te tapeke 10.

$$a=8 b=2$$

Tuhia anō te -16x^{2}+10x-1 hei \left(-16x^{2}+8x\right)+\left(2x-1\right).

$$\left(-16x^{2}+8x\right)+\left(2x-1\right)$$

Whakatauwehea atu -8x i te -16x^{2}+8x.

$$-8x\left(2x-1\right)+2x-1$$

Whakatauwehea atu te kīanga pātahi 2x-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2x-1\right)\left(-8x+1\right)$$

Hei kimi otinga whārite, me whakaoti te 2x-1=0 me te -8x+1=0.

$$x=\frac{1}{2} x=\frac{1}{8}$$

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.

$$-80x^{2}+50x-5=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -80 mō a, 50 mō b, me -5 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-50±\sqrt{50^{2}-4\left(-80\right)\left(-5\right)}}{2\left(-80\right)}$$

Pūrua 50.

$$x=\frac{-50±\sqrt{2500-4\left(-80\right)\left(-5\right)}}{2\left(-80\right)}$$

Whakareatia -4 ki te -80.

$$x=\frac{-50±\sqrt{2500+320\left(-5\right)}}{2\left(-80\right)}$$

Whakareatia 320 ki te -5.

$$x=\frac{-50±\sqrt{2500-1600}}{2\left(-80\right)}$$

Tāpiri 2500 ki te -1600.

$$x=\frac{-50±\sqrt{900}}{2\left(-80\right)}$$

Tuhia te pūtakerua o te 900.

$$x=\frac{-50±30}{2\left(-80\right)}$$

Whakareatia 2 ki te -80.

$$x=\frac{-50±30}{-160}$$

Nā, me whakaoti te whārite x=\frac{-50±30}{-160} ina he tāpiri te ±. Tāpiri -50 ki te 30.

$$x=-\frac{20}{-160}$$

Whakahekea te hautanga \frac{-20}{-160} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 20.

$$x=\frac{1}{8}$$

Nā, me whakaoti te whārite x=\frac{-50±30}{-160} ina he tango te ±. Tango 30 mai i -50.

$$x=-\frac{80}{-160}$$

Whakahekea te hautanga \frac{-80}{-160} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 80.

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

Kua oti te whārite te whakatau.

$$x=\frac{1}{8} x=\frac{1}{2}$$

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.

$$-80x^{2}+50x-5=0$$

Me tāpiri 5 ki ngā taha e rua o te whārite.

$$-80x^{2}+50x-5-\left(-5\right)=-\left(-5\right)$$

Mā te tango i te -5 i a ia ake anō ka toe ko te 0.

$$-80x^{2}+50x=-\left(-5\right)$$

Tango -5 mai i 0.

$$-80x^{2}+50x=5$$

Whakawehea ngā taha e rua ki te -80.

$$\frac{-80x^{2}+50x}{-80}=\frac{5}{-80}$$

Mā te whakawehe ki te -80 ka wetekia te whakareanga ki te -80.

$$x^{2}+\frac{50}{-80}x=\frac{5}{-80}$$

Whakahekea te hautanga \frac{50}{-80} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 10.

$$x^{2}-\frac{5}{8}x=\frac{5}{-80}$$

Whakahekea te hautanga \frac{5}{-80} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 5.

$$x^{2}-\frac{5}{8}x=-\frac{1}{16}$$

Whakawehea te -\frac{5}{8}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{5}{16}. Nā, tāpiria te pūrua o te -\frac{5}{16} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{5}{8}x+\left(-\frac{5}{16}\right)^{2}=-\frac{1}{16}+\left(-\frac{5}{16}\right)^{2}$$

Pūruatia -\frac{5}{16} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{5}{8}x+\frac{25}{256}=-\frac{1}{16}+\frac{25}{256}$$

Tāpiri -\frac{1}{16} ki te \frac{25}{256} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x^{2}-\frac{5}{8}x+\frac{25}{256}=\frac{9}{256}$$

Tauwehea x^{2}-\frac{5}{8}x+\frac{25}{256}. 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}.

$$\left(x-\frac{5}{16}\right)^{2}=\frac{9}{256}$$

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

$$\sqrt{\left(x-\frac{5}{16}\right)^{2}}=\sqrt{\frac{9}{256}}$$

Whakarūnātia.

$$x-\frac{5}{16}=\frac{3}{16} x-\frac{5}{16}=-\frac{3}{16}$$

Me tāpiri \frac{5}{16} ki ngā taha e rua o te whārite.

$$x=\frac{1}{2} x=\frac{1}{8}$$