Whakaoti i te left(x+2right)^2-9=-5

$${ \left(x+2 \right) }^{ 2 } -9=-5$$

Whakaoti mō x

$$x=0$$

$$x=-4$$

Ngā Upane Whakamahi Tauwehe

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+2\right)^{2}.

$$x^{2}+4x+4-9=-5$$

Tangohia te 9 i te 4, 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

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+2\right)^{2}.

$$x^{2}+4x+4-9=-5$$

Tangohia te 9 i te 4, 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.

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

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}.

$$x=\frac{-4±\sqrt{4^{2}}}{2}$$

Tuhia te pūtakerua o te 4^{2}.

$$x=\frac{-4±4}{2}$$

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

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

Whakawehe 0 ki te 2.

$$x=0$$

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

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

Whakawehe -8 ki te 2.

$$x=-4$$

Kua oti te whārite te whakatau.

$$x=0 x=-4$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+2\right)^{2}.

$$x^{2}+4x+4-9=-5$$

Tangohia te 9 i te 4, 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.

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

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 tango 2 mai i ngā taha e rua o te whārite.

$$x=0 x=-4$$