Whakaoti i te 4x^2-4x-15=0

$$4 { x }^{ 2 } -4x-15 = 0$$

Whakaoti mō x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

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 4x^{2}+ax+bx-15. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-4 ab=4\left(-15\right)=-60$$

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 -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Tātaihia te tapeke mō ia takirua.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

Ko te otinga te takirua ka hoatu i te tapeke -4.

$$a=-10 b=6$$

Tuhia anō te 4x^{2}-4x-15 hei \left(4x^{2}-10x\right)+\left(6x-15\right).

$$\left(4x^{2}-10x\right)+\left(6x-15\right)$$

Tauwehea te 2x i te tuatahi me te 3 i te rōpū tuarua.

$$2x\left(2x-5\right)+3\left(2x-5\right)$$

Whakatauwehea atu te kīanga pātahi 2x-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2x-5\right)\left(2x+3\right)$$

Hei kimi otinga whārite, me whakaoti te 2x-5=0 me te 2x+3=0.

$$x=\frac{5}{2} x=-\frac{3}{2}$$

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.

$$4x^{2}-4x-15=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 4 mō a, -4 mō b, me -15 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 4\left(-15\right)}}{2\times 4}$$

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 4\left(-15\right)}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-\left(-4\right)±\sqrt{16-16\left(-15\right)}}{2\times 4}$$

Whakareatia -16 ki te -15.

$$x=\frac{-\left(-4\right)±\sqrt{16+240}}{2\times 4}$$

Tāpiri 16 ki te 240.

$$x=\frac{-\left(-4\right)±\sqrt{256}}{2\times 4}$$

Tuhia te pūtakerua o te 256.

$$x=\frac{-\left(-4\right)±16}{2\times 4}$$

Ko te tauaro o -4 ko 4.

$$x=\frac{4±16}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{4±16}{8}$$

Nā, me whakaoti te whārite x=\frac{4±16}{8} ina he tāpiri te ±. Tāpiri 4 ki te 16.

$$x=\frac{20}{8}$$

Whakahekea te hautanga \frac{20}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$x=\frac{5}{2}$$

Nā, me whakaoti te whārite x=\frac{4±16}{8} ina he tango te ±. Tango 16 mai i 4.

$$x=-\frac{12}{8}$$

Whakahekea te hautanga \frac{-12}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

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

Kua oti te whārite te whakatau.

$$x=\frac{5}{2} x=-\frac{3}{2}$$

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.

$$4x^{2}-4x-15=0$$

Me tāpiri 15 ki ngā taha e rua o te whārite.

$$4x^{2}-4x-15-\left(-15\right)=-\left(-15\right)$$

Mā te tango i te -15 i a ia ake anō ka toe ko te 0.

$$4x^{2}-4x=-\left(-15\right)$$

Tango -15 mai i 0.

$$4x^{2}-4x=15$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4x^{2}-4x}{4}=\frac{15}{4}$$

Mā te whakawehe ki te 4 ka wetekia te whakareanga ki te 4.

$$x^{2}+\left(-\frac{4}{4}\right)x=\frac{15}{4}$$

Whakawehe -4 ki te 4.

$$x^{2}-x=\frac{15}{4}$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\frac{15}{4}+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=\frac{15+1}{4}$$

Tāpiri \frac{15}{4} ki te \frac{1}{4} 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.

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

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}.

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

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

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

Whakarūnātia.

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

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

$$x=\frac{5}{2} x=-\frac{3}{2}$$