Whakaoti i te (x+5)^2-36=0.u

$$( x + 5 ) ^ { 2 } - 36 = 0 . u$$

Whakaoti mō x

$$x=1$$

$$x=-11$$

Ngā Upane Whakamahi Tauwehe

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+5\right)^{2}.

$$x^{2}+10x+25-36=0$$

Tangohia te 36 i te 25, ka -11.

$$x^{2}+10x-11=0$$

Hei whakaoti i te whārite, whakatauwehea te x^{2}+10x-11 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.

$$a+b=10 ab=-11$$

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.

$$a=-1 b=11$$

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

$$\left(x-1\right)\left(x+11\right)$$

Hei kimi otinga whārite, me whakaoti te x-1=0 me te x+11=0.

$$x=1 x=-11$$

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+5\right)^{2}.

$$x^{2}+10x+25-36=0$$

Tangohia te 36 i te 25, ka -11.

$$x^{2}+10x-11=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx-11. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=10 ab=1\left(-11\right)=-11$$

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.

$$a=-1 b=11$$

Tuhia anō te x^{2}+10x-11 hei \left(x^{2}-x\right)+\left(11x-11\right).

$$\left(x^{2}-x\right)+\left(11x-11\right)$$

Tauwehea te x i te tuatahi me te 11 i te rōpū tuarua.

$$x\left(x-1\right)+11\left(x-1\right)$$

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

$$\left(x-1\right)\left(x+11\right)$$

Hei kimi otinga whārite, me whakaoti te x-1=0 me te x+11=0.

$$x=1 x=-11$$

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+5\right)^{2}.

$$x^{2}+10x+25-36=0$$

Tangohia te 36 i te 25, ka -11.

$$x^{2}+10x-11=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 10 mō b, me -11 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-11\right)}}{2}$$

Pūrua 10.

$$x=\frac{-10±\sqrt{100-4\left(-11\right)}}{2}$$

Whakareatia -4 ki te -11.

$$x=\frac{-10±\sqrt{100+44}}{2}$$

Tāpiri 100 ki te 44.

$$x=\frac{-10±\sqrt{144}}{2}$$

Tuhia te pūtakerua o te 144.

$$x=\frac{-10±12}{2}$$

Nā, me whakaoti te whārite x=\frac{-10±12}{2} ina he tāpiri te ±. Tāpiri -10 ki te 12.

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

Whakawehe 2 ki te 2.

$$x=1$$

Nā, me whakaoti te whārite x=\frac{-10±12}{2} ina he tango te ±. Tango 12 mai i -10.

$$x=-\frac{22}{2}$$

Whakawehe -22 ki te 2.

$$x=-11$$

Kua oti te whārite te whakatau.

$$x=1 x=-11$$

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(x+5\right)^{2}.

$$x^{2}+10x+25-36=0$$

Tangohia te 36 i te 25, ka -11.

$$x^{2}+10x-11=0$$

Me tāpiri te 11 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$x^{2}+10x=11$$

Whakawehea te 10, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te 5. Nā, tāpiria te pūrua o te 5 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+10x+5^{2}=11+5^{2}$$

Pūrua 5.

$$x^{2}+10x+25=11+25$$

Tāpiri 11 ki te 25.

$$x^{2}+10x+25=36$$

Tauwehea x^{2}+10x+25. 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}.

$$\left(x+5\right)^{2}=36$$

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

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{36}$$

Whakarūnātia.

$$x+5=6 x+5=-6$$

Me tango 5 mai i ngā taha e rua o te whārite.

$$x=1 x=-11$$