Whakaoti i te (2x+1)^2=sqrt{16}

$$( 2 x + 1 ) ^ { 2 } = \sqrt { 16 }$$

Whakaoti mō x

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

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

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(2x+1\right)^{2}.

$$4x^{2}+4x+1=\sqrt{16}$$

Tātaitia te pūtakerua o 16 kia tae ki 4.

$$4x^{2}+4x+1=4$$

Tangohia te 4 mai i ngā taha e rua.

$$4x^{2}+4x+1-4=0$$

Tangohia te 4 i te 1, ka -3.

$$4x^{2}+4x-3=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 4x^{2}+ax+bx-3. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=4 ab=4\left(-3\right)=-12$$

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. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -12.

$$-1,12 -2,6 -3,4$$

Tātaihia te tapeke mō ia takirua.

$$-1+12=11 -2+6=4 -3+4=1$$

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

$$a=-2 b=6$$

Tuhia anō te 4x^{2}+4x-3 hei \left(4x^{2}-2x\right)+\left(6x-3\right).

$$\left(4x^{2}-2x\right)+\left(6x-3\right)$$

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

$$2x\left(2x-1\right)+3\left(2x-1\right)$$

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

$$\left(2x-1\right)\left(2x+3\right)$$

Hei kimi otinga whārite, me whakaoti te 2x-1=0 me te 2x+3=0.

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(2x+1\right)^{2}.

$$4x^{2}+4x+1=\sqrt{16}$$

Tātaitia te pūtakerua o 16 kia tae ki 4.

$$4x^{2}+4x+1=4$$

Tangohia te 4 mai i ngā taha e rua.

$$4x^{2}+4x+1-4=0$$

Tangohia te 4 i te 1, ka -3.

$$4x^{2}+4x-3=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 4 mō a, 4 mō b, me -3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 4\left(-3\right)}}{2\times 4}$$

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\times 4\left(-3\right)}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-4±\sqrt{16-16\left(-3\right)}}{2\times 4}$$

Whakareatia -16 ki te -3.

$$x=\frac{-4±\sqrt{16+48}}{2\times 4}$$

Tāpiri 16 ki te 48.

$$x=\frac{-4±\sqrt{64}}{2\times 4}$$

Tuhia te pūtakerua o te 64.

$$x=\frac{-4±8}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-4±8}{8}$$

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

$$x=\frac{4}{8}$$

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

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

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

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

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(2x+1\right)^{2}.

$$4x^{2}+4x+1=\sqrt{16}$$

Tātaitia te pūtakerua o 16 kia tae ki 4.

$$4x^{2}+4x+1=4$$

Tangohia te 1 mai i ngā taha e rua.

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

Tangohia te 1 i te 4, ka 3.

$$4x^{2}+4x=3$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4x^{2}+4x}{4}=\frac{3}{4}$$

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

$$x^{2}+\frac{4}{4}x=\frac{3}{4}$$

Whakawehe 4 ki te 4.

$$x^{2}+x=\frac{3}{4}$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{3}{4}+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=\frac{3+1}{4}$$

Tāpiri \frac{3}{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.

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

Tauwehea te x^{2}+x+\frac{1}{4}. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

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

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

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

Whakarūnātia.

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

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

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