Whakaoti i te (x+5)^2=0

$$( x + 5 ) ^ { 2 } = 0$$

Whakaoti mō x

$$x=-5$$

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}.

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

Hei whakaoti i te whārite, whakatauwehea te x^{2}+10x+25 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.

$$a+b=10 ab=25$$

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 25.

$$1,25 5,5$$

Tātaihia te tapeke mō ia takirua.

$$1+25=26 5+5=10$$

Ko te otinga te takirua ka hoatu i te tapeke 10.

$$a=5 b=5$$

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

$$\left(x+5\right)\left(x+5\right)$$

Tuhia anōtia hei pūrua huarua.

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

Hei kimi i te otinga whārite, whakaotia te x+5=0.

$$x=-5$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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

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

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx+25. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=10 ab=1\times 25=25$$

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 25.

$$1,25 5,5$$

Tātaihia te tapeke mō ia takirua.

$$1+25=26 5+5=10$$

Ko te otinga te takirua ka hoatu i te tapeke 10.

$$a=5 b=5$$

Tuhia anō te x^{2}+10x+25 hei \left(x^{2}+5x\right)+\left(5x+25\right).

$$\left(x^{2}+5x\right)+\left(5x+25\right)$$

Tauwehea te x i te tuatahi me te 5 i te rōpū tuarua.

$$x\left(x+5\right)+5\left(x+5\right)$$

Whakatauwehea atu te kīanga pātahi x+5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x+5\right)\left(x+5\right)$$

Tuhia anōtia hei pūrua huarua.

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

Hei kimi i te otinga whārite, whakaotia te x+5=0.

$$x=-5$$

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}.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 10 mō b, me 25 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 25}}{2}$$

Pūrua 10.

$$x=\frac{-10±\sqrt{100-4\times 25}}{2}$$

Whakareatia -4 ki te 25.

$$x=\frac{-10±\sqrt{100-100}}{2}$$

Tāpiri 100 ki te -100.

$$x=\frac{-10±\sqrt{0}}{2}$$

Tuhia te pūtakerua o te 0.

$$x=-\frac{10}{2}$$

Whakawehe -10 ki te 2.

$$x=-5$$

Ngā Upane mō te Whakaoti i te Pūrua

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

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{0}$$

Whakarūnātia.

$$x+5=0 x+5=0$$

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

$$x=-5 x=-5$$

Kua oti te whārite te whakatau. He ōrite ngā whakatau.

$$x=-5$$