Whakaoti i te (x+3)^2=16

$$( x + 3 ) ^ { 2 } = 16$$

Whakaoti mō x

$$x=1$$

$$x=-7$$

Ngā Upane Whakamahi Tauwehe

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+3\right)^{2}.

$$x^{2}+6x+9=16$$

Tangohia te 16 mai i ngā taha e rua.

$$x^{2}+6x+9-16=0$$

Tangohia te 16 i te 9, ka -7.

$$x^{2}+6x-7=0$$

Hei whakaoti i te whārite, whakatauwehea te x^{2}+6x-7 mā te whakamahi i te tātai x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=6 ab=-7$$

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.

$$a=-1 b=7$$

Me tuhi anō te kīanga whakatauwehe \left(x+a\right)\left(x+b\right) mā ngā uara i tātaihia.

$$\left(x-1\right)\left(x+7\right)$$

Hei kimi otinga whārite, me whakaoti te x-1=0 me te x+7=0.

$$x=1 x=-7$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+3\right)^{2}.

$$x^{2}+6x+9=16$$

Tangohia te 16 mai i ngā taha e rua.

$$x^{2}+6x+9-16=0$$

Tangohia te 16 i te 9, ka -7.

$$x^{2}+6x-7=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx-7. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=6 ab=1\left(-7\right)=-7$$

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.

$$a=-1 b=7$$

Tuhia anō te x^{2}+6x-7 hei \left(x^{2}-x\right)+\left(7x-7\right).

$$\left(x^{2}-x\right)+\left(7x-7\right)$$

Tauwehea te x i te tuatahi me te 7 i te rōpū tuarua.

$$x\left(x-1\right)+7\left(x-1\right)$$

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

$$\left(x-1\right)\left(x+7\right)$$

Hei kimi otinga whārite, me whakaoti te x-1=0 me te x+7=0.

$$x=1 x=-7$$

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(x+3\right)^{2}.

$$x^{2}+6x+9=16$$

Tangohia te 16 mai i ngā taha e rua.

$$x^{2}+6x+9-16=0$$

Tangohia te 16 i te 9, ka -7.

$$x^{2}+6x-7=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 6 mō b, me -7 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-7\right)}}{2}$$

Pūrua 6.

$$x=\frac{-6±\sqrt{36-4\left(-7\right)}}{2}$$

Whakareatia -4 ki te -7.

$$x=\frac{-6±\sqrt{36+28}}{2}$$

Tāpiri 36 ki te 28.

$$x=\frac{-6±\sqrt{64}}{2}$$

Tuhia te pūtakerua o te 64.

$$x=\frac{-6±8}{2}$$

Nā, me whakaoti te whārite x=\frac{-6±8}{2} ina he tāpiri te ±. Tāpiri -6 ki te 8.

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

Whakawehe 2 ki te 2.

$$x=1$$

Nā, me whakaoti te whārite x=\frac{-6±8}{2} ina he tango te ±. Tango 8 mai i -6.

$$x=-\frac{14}{2}$$

Whakawehe -14 ki te 2.

$$x=-7$$

Kua oti te whārite te whakatau.

$$x=1 x=-7$$

Ngā Upane mō te Whakaoti i te Pūrua

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

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{16}$$

Whakarūnātia.

$$x+3=4 x+3=-4$$

Me tango 3 mai i ngā taha e rua o te whārite.

$$x=1 x=-7$$