Whakaoti i te left(x+1right)^2=1-3x

$${ \left(x+1 \right) }^{ 2 } = 1-3x$$

Whakaoti mō x

$$x=-5$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+1\right)^{2}.

$$x^{2}+2x+1=1-3x$$

Tangohia te 1 mai i ngā taha e rua.

$$x^{2}+2x+1-1=-3x$$

Tangohia te 1 i te 1, ka 0.

$$x^{2}+2x=-3x$$

Me tāpiri te 3x ki ngā taha e rua.

$$x^{2}+2x+3x=0$$

Pahekotia te 2x me 3x, ka 5x.

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

Tauwehea te x.

$$x\left(x+5\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te x+5=0.

$$x=0 x=-5$$

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+1\right)^{2}.

$$x^{2}+2x+1=1-3x$$

Tangohia te 1 mai i ngā taha e rua.

$$x^{2}+2x+1-1=-3x$$

Tangohia te 1 i te 1, ka 0.

$$x^{2}+2x=-3x$$

Me tāpiri te 3x ki ngā taha e rua.

$$x^{2}+2x+3x=0$$

Pahekotia te 2x me 3x, ka 5x.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 5 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}}}{2}$$

Tuhia te pūtakerua o te 5^{2}.

$$x=\frac{-5±5}{2}$$

Nā, me whakaoti te whārite x=\frac{-5±5}{2} ina he tāpiri te ±. Tāpiri -5 ki te 5.

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

Whakawehe 0 ki te 2.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-5±5}{2} ina he tango te ±. Tango 5 mai i -5.

$$x=-\frac{10}{2}$$

Whakawehe -10 ki te 2.

$$x=-5$$

Kua oti te whārite te whakatau.

$$x=0 x=-5$$

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+1\right)^{2}.

$$x^{2}+2x+1=1-3x$$

Tangohia te 1 mai i ngā taha e rua.

$$x^{2}+2x+1-1=-3x$$

Tangohia te 1 i te 1, ka 0.

$$x^{2}+2x=-3x$$

Me tāpiri te 3x ki ngā taha e rua.

$$x^{2}+2x+3x=0$$

Pahekotia te 2x me 3x, ka 5x.

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

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 tango \frac{5}{2} mai i ngā taha e rua o te whārite.

$$x=0 x=-5$$