Whakaoti i te 3x(x-5)=0

$$3x(x-5)=0$$

Whakaoti mō x

$$x=5$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Whakamahia te āhuatanga tohatoha hei whakarea te 3x ki te x-5.

$$3x^{2}-15x=0$$

Tauwehea te x.

$$x\left(3x-15\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 3x-15=0.

$$x=0 x=5$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 3x ki te x-5.

$$3x^{2}-15x=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, -15 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}}}{2\times 3}$$

Tuhia te pūtakerua o te \left(-15\right)^{2}.

$$x=\frac{-\left(-15\right)±15}{2\times 3}$$

Ko te tauaro o -15 ko 15.

$$x=\frac{15±15}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{15±15}{6}$$

Nā, me whakaoti te whārite x=\frac{15±15}{6} ina he tāpiri te ±. Tāpiri 15 ki te 15.

$$x=\frac{30}{6}$$

Whakawehe 30 ki te 6.

$$x=5$$

Nā, me whakaoti te whārite x=\frac{15±15}{6} ina he tango te ±. Tango 15 mai i 15.

$$x=\frac{0}{6}$$

Whakawehe 0 ki te 6.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=5 x=0$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 3x ki te x-5.

$$3x^{2}-15x=0$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{3x^{2}-15x}{3}=\frac{0}{3}$$

Mā te whakawehe ki te 3 ka wetekia te whakareanga ki te 3.

$$x^{2}+\left(-\frac{15}{3}\right)x=\frac{0}{3}$$

Whakawehe -15 ki te 3.

$$x^{2}-5x=\frac{0}{3}$$

Whakawehe 0 ki te 3.

$$x^{2}-5x=0$$

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

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

Pūruatia -\frac{5}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-5x+\frac{25}{4}=\frac{25}{4}$$

Tauwehea x^{2}-5x+\frac{25}{4}. 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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{25}{4}$$

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

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Whakarūnātia.

$$x-\frac{5}{2}=\frac{5}{2} x-\frac{5}{2}=-\frac{5}{2}$$

Me tāpiri \frac{5}{2} ki ngā taha e rua o te whārite.

$$x=5 x=0$$