Whakaoti i te x^2-4x-5=0-5

$${ x }^{ 2 } -4x-5 = 0-5$$

Whakaoti mō x

$$x=4$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Tangohia te 5 i te 0, ka -5.

$$x^{2}-4x-5=-5$$

Me tāpiri te 5 ki ngā taha e rua.

$$x^{2}-4x-5+5=0$$

Tāpirihia te -5 ki te 5, ka 0.

$$x^{2}-4x=0$$

Tauwehea te x.

$$x\left(x-4\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te x-4=0.

$$x=0 x=4$$

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.

$$x^{2}-4x-5=-5$$

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

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

Mā te tango i te -5 i a ia ake anō ka toe ko te 0.

$$x^{2}-4x-5-\left(-5\right)=0$$

Tango -5 mai i -5.

$$x^{2}-4x=0$$

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

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

Tuhia te pūtakerua o te \left(-4\right)^{2}.

$$x=\frac{-\left(-4\right)±4}{2}$$

Ko te tauaro o -4 ko 4.

$$x=\frac{4±4}{2}$$

Nā, me whakaoti te whārite x=\frac{4±4}{2} ina he tāpiri te ±. Tāpiri 4 ki te 4.

$$x=\frac{8}{2}$$

Whakawehe 8 ki te 2.

$$x=4$$

Nā, me whakaoti te whārite x=\frac{4±4}{2} ina he tango te ±. Tango 4 mai i 4.

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

Whakawehe 0 ki te 2.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=4 x=0$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 5 i te 0, ka -5.

$$x^{2}-4x-5=-5$$

Me tāpiri te 5 ki ngā taha e rua.

$$x^{2}-4x-5+5=0$$

Tāpirihia te -5 ki te 5, ka 0.

$$x^{2}-4x=0$$

Whakawehea te -4, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -2. Nā, tāpiria te pūrua o te -2 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-4x+\left(-2\right)^{2}=\left(-2\right)^{2}$$

Pūrua -2.

$$x^{2}-4x+4=4$$

Tauwehea x^{2}-4x+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}.

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

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

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

Whakarūnātia.

$$x-2=2 x-2=-2$$

Me tāpiri 2 ki ngā taha e rua o te whārite.

$$x=4 x=0$$