Whakaoti i te 5x=x^2+25/2

$$5 x = \frac { x ^ { 2 } + 25 } { 2 }$$

Whakaoti mō x

$$x=5$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakareatia ngā taha e rua o te whārite ki te 2.

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

Tangohia te x^{2} mai i ngā taha e rua.

$$10x-x^{2}=25$$

Tangohia te 25 mai i ngā taha e rua.

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

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

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

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.

$$a+b=10 ab=-\left(-25\right)=25$$

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.

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

Hei kimi otinga whārite, me whakaoti te x-5=0 me te -x+5=0.

$$x=5 x=5$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia ngā taha e rua o te whārite ki te 2.

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

Tangohia te x^{2} mai i ngā taha e rua.

$$10x-x^{2}=25$$

Tangohia te 25 mai i ngā taha e rua.

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

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

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

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}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-1\right)\left(-25\right)}}{2\left(-1\right)}$$

Pūrua 10.

$$x=\frac{-10±\sqrt{100-4\left(-1\right)\left(-25\right)}}{2\left(-1\right)}$$

Whakareatia -4 ki te -1.

$$x=\frac{-10±\sqrt{100+4\left(-25\right)}}{2\left(-1\right)}$$

Whakareatia 4 ki te -25.

$$x=\frac{-10±\sqrt{100-100}}{2\left(-1\right)}$$

Tāpiri 100 ki te -100.

$$x=\frac{-10±\sqrt{0}}{2\left(-1\right)}$$

Tuhia te pūtakerua o te 0.

$$x=-\frac{10}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

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

Whakawehe -10 ki te -2.

$$x=5$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakareatia ngā taha e rua o te whārite ki te 2.

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

Tangohia te x^{2} mai i ngā taha e rua.

$$10x-x^{2}=25$$

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

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

Whakawehea ngā taha e rua ki te -1.

$$\frac{-x^{2}+10x}{-1}=\frac{25}{-1}$$

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

$$x^{2}+\frac{10}{-1}x=\frac{25}{-1}$$

Whakawehe 10 ki te -1.

$$x^{2}-10x=\frac{25}{-1}$$

Whakawehe 25 ki te -1.

$$x^{2}-10x=-25$$

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.

$$x^{2}-10x+\left(-5\right)^{2}=-25+\left(-5\right)^{2}$$

Pūrua -5.

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

Tāpiri -25 ki te 25.

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

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}.

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

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 tāpiri 5 ki ngā taha e rua o te whārite.

$$x=5 x=5$$

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

$$x=5$$