Whakaoti i te 50+x^2-10x=50

$$50 + x ^ { 2 } - 10 x = 50$$

Whakaoti mō x

$$x=10$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Tangohia te 50 mai i ngā taha e rua.

$$50+x^{2}-10x-50=0$$

Tangohia te 50 i te 50, ka 0.

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

Tauwehea te x.

$$x\left(x-10\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te x-10=0.

$$x=0 x=10$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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.

$$x^{2}-10x+50=50$$

Me tango 50 mai i ngā taha e rua o te whārite.

$$x^{2}-10x+50-50=50-50$$

Mā te tango i te 50 i a ia ake anō ka toe ko te 0.

$$x^{2}-10x+50-50=0$$

Tango 50 mai i 50.

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

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

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

Tuhia te pūtakerua o te \left(-10\right)^{2}.

$$x=\frac{-\left(-10\right)±10}{2}$$

Ko te tauaro o -10 ko 10.

$$x=\frac{10±10}{2}$$

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

$$x=\frac{20}{2}$$

Whakawehe 20 ki te 2.

$$x=10$$

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

$$x=\frac{0}{2}$$

Whakawehe 0 ki te 2.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=10 x=0$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 50 mai i ngā taha e rua.

$$50+x^{2}-10x-50=0$$

Tangohia te 50 i te 50, ka 0.

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

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.

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

Pūrua -5.

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

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}.

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

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

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

Whakarūnātia.

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

Me tāpiri 5 ki ngā taha e rua o te whārite.

$$x=10 x=0$$