Whakaoti i te (x+3)(x+1)=2x+7

$$( x + 3 ) ( x + 1 ) = 2 x + 7$$

Whakaoti mō x (complex solution)

$$x=\sqrt{5}-1\approx 1.236067977$$

$$x=-\left(\sqrt{5}+1\right)\approx -3.236067977$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te x+3 ki te x+1 ka whakakotahi i ngā kupu rite.

$$x^{2}+4x+3=2x+7$$

Tangohia te 2x mai i ngā taha e rua.

$$x^{2}+4x+3-2x=7$$

Pahekotia te 4x me -2x, ka 2x.

$$x^{2}+2x+3=7$$

Tangohia te 7 mai i ngā taha e rua.

$$x^{2}+2x+3-7=0$$

Tangohia te 7 i te 3, ka -4.

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

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

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

Pūrua 2.

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

Whakareatia -4 ki te -4.

$$x=\frac{-2±\sqrt{4+16}}{2}$$

Tāpiri 4 ki te 16.

$$x=\frac{-2±\sqrt{20}}{2}$$

Tuhia te pūtakerua o te 20.

$$x=\frac{-2±2\sqrt{5}}{2}$$

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

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

Whakawehe -2+2\sqrt{5} ki te 2.

$$x=\sqrt{5}-1$$

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

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

Whakawehe -2-2\sqrt{5} ki te 2.

$$x=-\sqrt{5}-1$$

Kua oti te whārite te whakatau.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te x+3 ki te x+1 ka whakakotahi i ngā kupu rite.

$$x^{2}+4x+3=2x+7$$

Tangohia te 2x mai i ngā taha e rua.

$$x^{2}+4x+3-2x=7$$

Pahekotia te 4x me -2x, ka 2x.

$$x^{2}+2x+3=7$$

Tangohia te 3 mai i ngā taha e rua.

$$x^{2}+2x=7-3$$

Tangohia te 3 i te 7, ka 4.

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

Whakawehea te 2, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te 1. Nā, tāpiria te pūrua o te 1 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+2x+1^{2}=4+1^{2}$$

Pūrua 1.

$$x^{2}+2x+1=4+1$$

Tāpiri 4 ki te 1.

$$x^{2}+2x+1=5$$

Tauwehea x^{2}+2x+1. 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}.

$$\left(x+1\right)^{2}=5$$

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

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{5}$$

Whakarūnātia.

$$x+1=\sqrt{5} x+1=-\sqrt{5}$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$

Whakaoti mō x

$$x=\sqrt{5}-1\approx 1.236067977$$

$$x=-\sqrt{5}-1\approx -3.236067977$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te x+3 ki te x+1 ka whakakotahi i ngā kupu rite.

$$x^{2}+4x+3=2x+7$$

Tangohia te 2x mai i ngā taha e rua.

$$x^{2}+4x+3-2x=7$$

Pahekotia te 4x me -2x, ka 2x.

$$x^{2}+2x+3=7$$

Tangohia te 7 mai i ngā taha e rua.

$$x^{2}+2x+3-7=0$$

Tangohia te 7 i te 3, ka -4.

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

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

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

Pūrua 2.

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

Whakareatia -4 ki te -4.

$$x=\frac{-2±\sqrt{4+16}}{2}$$

Tāpiri 4 ki te 16.

$$x=\frac{-2±\sqrt{20}}{2}$$

Tuhia te pūtakerua o te 20.

$$x=\frac{-2±2\sqrt{5}}{2}$$

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

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

Whakawehe -2+2\sqrt{5} ki te 2.

$$x=\sqrt{5}-1$$

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

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

Whakawehe -2-2\sqrt{5} ki te 2.

$$x=-\sqrt{5}-1$$

Kua oti te whārite te whakatau.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te x+3 ki te x+1 ka whakakotahi i ngā kupu rite.

$$x^{2}+4x+3=2x+7$$

Tangohia te 2x mai i ngā taha e rua.

$$x^{2}+4x+3-2x=7$$

Pahekotia te 4x me -2x, ka 2x.

$$x^{2}+2x+3=7$$

Tangohia te 3 mai i ngā taha e rua.

$$x^{2}+2x=7-3$$

Tangohia te 3 i te 7, ka 4.

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

Whakawehea te 2, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te 1. Nā, tāpiria te pūrua o te 1 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+2x+1^{2}=4+1^{2}$$

Pūrua 1.

$$x^{2}+2x+1=4+1$$

Tāpiri 4 ki te 1.

$$x^{2}+2x+1=5$$

Tauwehea x^{2}+2x+1. 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}.

$$\left(x+1\right)^{2}=5$$

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

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{5}$$

Whakarūnātia.

$$x+1=\sqrt{5} x+1=-\sqrt{5}$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$