Whakaoti i te x^2+51/2x+6=0

$$x ^ { 2 } + 5 \frac { 1 } { 2 } x + 6 = 0$$

Whakaoti mō x

$$x=-4$$

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

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakareatia ngā taha e rua o te whārite ki te 2.

$$2x^{2}+\left(5\times 2+1\right)x+12=0$$

Whakareatia te 5 ki te 2, ka 10.

$$2x^{2}+\left(10+1\right)x+12=0$$

Tāpirihia te 10 ki te 1, ka 11.

$$2x^{2}+11x+12=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 2x^{2}+ax+bx+12. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=11 ab=2\times 12=24$$

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 24.

$$1,24 2,12 3,8 4,6$$

Tātaihia te tapeke mō ia takirua.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

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

$$a=3 b=8$$

Tuhia anō te 2x^{2}+11x+12 hei \left(2x^{2}+3x\right)+\left(8x+12\right).

$$\left(2x^{2}+3x\right)+\left(8x+12\right)$$

Tauwehea te x i te tuatahi me te 4 i te rōpū tuarua.

$$x\left(2x+3\right)+4\left(2x+3\right)$$

Whakatauwehea atu te kīanga pātahi 2x+3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2x+3\right)\left(x+4\right)$$

Hei kimi otinga whārite, me whakaoti te 2x+3=0 me te x+4=0.

$$x=-\frac{3}{2} x=-4$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia ngā taha e rua o te whārite ki te 2.

$$2x^{2}+\left(5\times 2+1\right)x+12=0$$

Whakareatia te 5 ki te 2, ka 10.

$$2x^{2}+\left(10+1\right)x+12=0$$

Tāpirihia te 10 ki te 1, ka 11.

$$2x^{2}+11x+12=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, 11 mō b, me 12 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\times 2\times 12}}{2\times 2}$$

Pūrua 11.

$$x=\frac{-11±\sqrt{121-4\times 2\times 12}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-11±\sqrt{121-8\times 12}}{2\times 2}$$

Whakareatia -8 ki te 12.

$$x=\frac{-11±\sqrt{121-96}}{2\times 2}$$

Tāpiri 121 ki te -96.

$$x=\frac{-11±\sqrt{25}}{2\times 2}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-11±5}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-11±5}{4}$$

Nā, me whakaoti te whārite x=\frac{-11±5}{4} ina he tāpiri te ±. Tāpiri -11 ki te 5.

$$x=-\frac{6}{4}$$

Whakahekea te hautanga \frac{-6}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Nā, me whakaoti te whārite x=\frac{-11±5}{4} ina he tango te ±. Tango 5 mai i -11.

$$x=-\frac{16}{4}$$

Whakawehe -16 ki te 4.

$$x=-4$$

Kua oti te whārite te whakatau.

$$x=-\frac{3}{2} x=-4$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakareatia ngā taha e rua o te whārite ki te 2.

$$2x^{2}+\left(5\times 2+1\right)x+12=0$$

Whakareatia te 5 ki te 2, ka 10.

$$2x^{2}+\left(10+1\right)x+12=0$$

Tāpirihia te 10 ki te 1, ka 11.

$$2x^{2}+11x+12=0$$

Tangohia te 12 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$2x^{2}+11x=-12$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}+11x}{2}=-\frac{12}{2}$$

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

$$x^{2}+\frac{11}{2}x=-\frac{12}{2}$$

Whakawehe -12 ki te 2.

$$x^{2}+\frac{11}{2}x=-6$$

Whakawehea te \frac{11}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{11}{4}. Nā, tāpiria te pūrua o te \frac{11}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{11}{2}x+\left(\frac{11}{4}\right)^{2}=-6+\left(\frac{11}{4}\right)^{2}$$

Pūruatia \frac{11}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{11}{2}x+\frac{121}{16}=-6+\frac{121}{16}$$

Tāpiri -6 ki te \frac{121}{16}.

$$x^{2}+\frac{11}{2}x+\frac{121}{16}=\frac{25}{16}$$

Tauwehea x^{2}+\frac{11}{2}x+\frac{121}{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}.

$$\left(x+\frac{11}{4}\right)^{2}=\frac{25}{16}$$

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

$$\sqrt{\left(x+\frac{11}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Whakarūnātia.

$$x+\frac{11}{4}=\frac{5}{4} x+\frac{11}{4}=-\frac{5}{4}$$

Me tango \frac{11}{4} mai i ngā taha e rua o te whārite.

$$x=-\frac{3}{2} x=-4$$