Whakaoti i te 25(1+2a+a^2)=14

$$25 ( 1 + 2 a + a ^ { 2 } ) = 14$$

Whakaoti mō a

$$a=\frac{\sqrt{14}}{5}-1\approx -0.251668523$$

$$a=-\frac{\sqrt{14}}{5}-1\approx -1.748331477$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakawehea ngā taha e rua ki te 25.

$$1+2a+a^{2}=\frac{14}{25}$$

Tangohia te \frac{14}{25} mai i ngā taha e rua.

$$1+2a+a^{2}-\frac{14}{25}=0$$

Tangohia te \frac{14}{25} i te 1, ka \frac{11}{25}.

$$\frac{11}{25}+2a+a^{2}=0$$

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.

$$a^{2}+2a+\frac{11}{25}=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 2 mō b, me \frac{11}{25} mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-2±\sqrt{2^{2}-4\times \frac{11}{25}}}{2}$$

Pūrua 2.

$$a=\frac{-2±\sqrt{4-4\times \frac{11}{25}}}{2}$$

Whakareatia -4 ki te \frac{11}{25}.

$$a=\frac{-2±\sqrt{4-\frac{44}{25}}}{2}$$

Tāpiri 4 ki te -\frac{44}{25}.

$$a=\frac{-2±\sqrt{\frac{56}{25}}}{2}$$

Tuhia te pūtakerua o te \frac{56}{25}.

$$a=\frac{-2±\frac{2\sqrt{14}}{5}}{2}$$

Nā, me whakaoti te whārite a=\frac{-2±\frac{2\sqrt{14}}{5}}{2} ina he tāpiri te ±. Tāpiri -2 ki te \frac{2\sqrt{14}}{5}.

$$a=\frac{\frac{2\sqrt{14}}{5}-2}{2}$$

Whakawehe -2+\frac{2\sqrt{14}}{5} ki te 2.

$$a=\frac{\sqrt{14}}{5}-1$$

Nā, me whakaoti te whārite a=\frac{-2±\frac{2\sqrt{14}}{5}}{2} ina he tango te ±. Tango \frac{2\sqrt{14}}{5} mai i -2.

$$a=\frac{-\frac{2\sqrt{14}}{5}-2}{2}$$

Whakawehe -2-\frac{2\sqrt{14}}{5} ki te 2.

$$a=-\frac{\sqrt{14}}{5}-1$$

Kua oti te whārite te whakatau.

$$a=\frac{\sqrt{14}}{5}-1 a=-\frac{\sqrt{14}}{5}-1$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakawehea ngā taha e rua ki te 25.

$$1+2a+a^{2}=\frac{14}{25}$$

Tangohia te 1 mai i ngā taha e rua.

$$2a+a^{2}=\frac{14}{25}-1$$

Tangohia te 1 i te \frac{14}{25}, ka -\frac{11}{25}.

$$2a+a^{2}=-\frac{11}{25}$$

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

$$a^{2}+2a=-\frac{11}{25}$$

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.

$$a^{2}+2a+1^{2}=-\frac{11}{25}+1^{2}$$

Pūrua 1.

$$a^{2}+2a+1=-\frac{11}{25}+1$$

Tāpiri -\frac{11}{25} ki te 1.

$$a^{2}+2a+1=\frac{14}{25}$$

Tauwehea a^{2}+2a+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}.

$$\left(a+1\right)^{2}=\frac{14}{25}$$

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

$$\sqrt{\left(a+1\right)^{2}}=\sqrt{\frac{14}{25}}$$

Whakarūnātia.

$$a+1=\frac{\sqrt{14}}{5} a+1=-\frac{\sqrt{14}}{5}$$

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

$$a=\frac{\sqrt{14}}{5}-1 a=-\frac{\sqrt{14}}{5}-1$$