Whakaoti i te 4x^2+8x=4x+8

$$4 { x }^{ 2 } +8x = 4x+8$$

Whakaoti mō x

$$x=-2$$

$$x=1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tangohia te 4x mai i ngā taha e rua.

$$4x^{2}+8x-4x=8$$

Pahekotia te 8x me -4x, ka 4x.

$$4x^{2}+4x=8$$

Tangohia te 8 mai i ngā taha e rua.

$$4x^{2}+4x-8=0$$

Whakawehea ngā taha e rua ki te 4.

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

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 tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-1 b=2$$

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

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

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

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

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

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

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

$$x=1 x=-2$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 4x mai i ngā taha e rua.

$$4x^{2}+8x-4x=8$$

Pahekotia te 8x me -4x, ka 4x.

$$4x^{2}+4x=8$$

Tangohia te 8 mai i ngā taha e rua.

$$4x^{2}+4x-8=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 4 mō a, 4 mō b, me -8 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 4\left(-8\right)}}{2\times 4}$$

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\times 4\left(-8\right)}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-4±\sqrt{16-16\left(-8\right)}}{2\times 4}$$

Whakareatia -16 ki te -8.

$$x=\frac{-4±\sqrt{16+128}}{2\times 4}$$

Tāpiri 16 ki te 128.

$$x=\frac{-4±\sqrt{144}}{2\times 4}$$

Tuhia te pūtakerua o te 144.

$$x=\frac{-4±12}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-4±12}{8}$$

Nā, me whakaoti te whārite x=\frac{-4±12}{8} ina he tāpiri te ±. Tāpiri -4 ki te 12.

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

Whakawehe 8 ki te 8.

$$x=1$$

Nā, me whakaoti te whārite x=\frac{-4±12}{8} ina he tango te ±. Tango 12 mai i -4.

$$x=-\frac{16}{8}$$

Whakawehe -16 ki te 8.

$$x=-2$$

Kua oti te whārite te whakatau.

$$x=1 x=-2$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 4x mai i ngā taha e rua.

$$4x^{2}+8x-4x=8$$

Pahekotia te 8x me -4x, ka 4x.

$$4x^{2}+4x=8$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4x^{2}+4x}{4}=\frac{8}{4}$$

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

$$x^{2}+\frac{4}{4}x=\frac{8}{4}$$

Whakawehe 4 ki te 4.

$$x^{2}+x=\frac{8}{4}$$

Whakawehe 8 ki te 4.

$$x^{2}+x=2$$

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 tango \frac{1}{2} mai i ngā taha e rua o te whārite.

$$x=1 x=-2$$