Whakaoti i te (5+a)^2+a=8+a

$$( 5 + a ) ^ { 2 } + a = 8 + a$$

Whakaoti mō a

$$a=2\sqrt{2}-5\approx -2.171572875$$

$$a=-2\sqrt{2}-5\approx -7.828427125$$

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(5+a\right)^{2}.

$$25+10a+a^{2}+a=8+a$$

Pahekotia te 10a me a, ka 11a.

$$25+11a+a^{2}=8+a$$

Tangohia te 8 mai i ngā taha e rua.

$$25+11a+a^{2}-8=a$$

Tangohia te 8 i te 25, ka 17.

$$17+11a+a^{2}=a$$

Tangohia te a mai i ngā taha e rua.

$$17+11a+a^{2}-a=0$$

Pahekotia te 11a me -a, ka 10a.

$$17+10a+a^{2}=0$$

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.

$$a^{2}+10a+17=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 10 mō b, me 17 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-10±\sqrt{10^{2}-4\times 17}}{2}$$

Pūrua 10.

$$a=\frac{-10±\sqrt{100-4\times 17}}{2}$$

Whakareatia -4 ki te 17.

$$a=\frac{-10±\sqrt{100-68}}{2}$$

Tāpiri 100 ki te -68.

$$a=\frac{-10±\sqrt{32}}{2}$$

Tuhia te pūtakerua o te 32.

$$a=\frac{-10±4\sqrt{2}}{2}$$

Nā, me whakaoti te whārite a=\frac{-10±4\sqrt{2}}{2} ina he tāpiri te ±. Tāpiri -10 ki te 4\sqrt{2}.

$$a=\frac{4\sqrt{2}-10}{2}$$

Whakawehe -10+4\sqrt{2} ki te 2.

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

Nā, me whakaoti te whārite a=\frac{-10±4\sqrt{2}}{2} ina he tango te ±. Tango 4\sqrt{2} mai i -10.

$$a=\frac{-4\sqrt{2}-10}{2}$$

Whakawehe -10-4\sqrt{2} ki te 2.

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

Kua oti te whārite te whakatau.

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

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(5+a\right)^{2}.

$$25+10a+a^{2}+a=8+a$$

Pahekotia te 10a me a, ka 11a.

$$25+11a+a^{2}=8+a$$

Tangohia te a mai i ngā taha e rua.

$$25+11a+a^{2}-a=8$$

Pahekotia te 11a me -a, ka 10a.

$$25+10a+a^{2}=8$$

Tangohia te 25 mai i ngā taha e rua.

$$10a+a^{2}=8-25$$

Tangohia te 25 i te 8, ka -17.

$$10a+a^{2}=-17$$

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.

$$a^{2}+10a=-17$$

Whakawehea te 10, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te 5. Nā, tāpiria te pūrua o te 5 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$a^{2}+10a+5^{2}=-17+5^{2}$$

Pūrua 5.

$$a^{2}+10a+25=-17+25$$

Tāpiri -17 ki te 25.

$$a^{2}+10a+25=8$$

Tauwehea a^{2}+10a+25. 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}.

$$\left(a+5\right)^{2}=8$$

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

$$\sqrt{\left(a+5\right)^{2}}=\sqrt{8}$$

Whakarūnātia.

$$a+5=2\sqrt{2} a+5=-2\sqrt{2}$$

Me tango 5 mai i ngā taha e rua o te whārite.

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