Whakaoti i te sqrt{x}=x/3

$$\sqrt{ x } = \frac{ x }{ 3 }$$

Whakaoti mō x

$$x=0$$

$$x=9$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

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

Tātaihia te \sqrt{x} mā te pū o 2, kia riro ko x.

$$x=\left(\frac{x}{3}\right)^{2}$$

Kia whakarewa i te \frac{x}{3} ki tētahi taupū, me whakarewa tahi te taurunga me te tauraro ki te taupū kātahi ka whakawehe.

$$x=\frac{x^{2}}{3^{2}}$$

Tātaihia te 3 mā te pū o 2, kia riro ko 9.

$$x=\frac{x^{2}}{9}$$

Tangohia te \frac{x^{2}}{9} mai i ngā taha e rua.

$$x-\frac{x^{2}}{9}=0$$

Whakareatia ngā taha e rua o te whārite ki te 9.

$$9x-x^{2}=0$$

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.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -1 mō a, 9 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}}}{2\left(-1\right)}$$

Tuhia te pūtakerua o te 9^{2}.

$$x=\frac{-9±9}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{-9±9}{-2}$$

Nā, me whakaoti te whārite x=\frac{-9±9}{-2} ina he tāpiri te ±. Tāpiri -9 ki te 9.

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

Whakawehe 0 ki te -2.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-9±9}{-2} ina he tango te ±. Tango 9 mai i -9.

$$x=-\frac{18}{-2}$$

Whakawehe -18 ki te -2.

$$x=9$$

Kua oti te whārite te whakatau.

$$x=0 x=9$$

Whakakapia te 0 mō te x i te whārite \sqrt{x}=\frac{x}{3}.

$$\sqrt{0}=\frac{0}{3}$$

Whakarūnātia. Ko te uara x=0 kua ngata te whārite.

$$0=0$$

Whakakapia te 9 mō te x i te whārite \sqrt{x}=\frac{x}{3}.

$$\sqrt{9}=\frac{9}{3}$$

Whakarūnātia. Ko te uara x=9 kua ngata te whārite.

$$3=3$$

Rārangihia ngā rongoā katoa o \sqrt{x}=\frac{x}{3}.

$$x=0 x=9$$