Whakaoti i te (a)=9a^2+46a.text{solvefor}h(a)=-5

$$( a ) = 9 a ^ { 2 } + 46 a . \text { solve for } h ( a ) = - 5$$

Whakaoti mō a

$$a=-5$$

$$a=0$$

Ngā Upane Whakamahi Tauwehe

Tangohia te 9a^{2} mai i ngā taha e rua.

$$a-9a^{2}=46a$$

Tangohia te 46a mai i ngā taha e rua.

$$a-9a^{2}-46a=0$$

Pahekotia te a me -46a, ka -45a.

$$-45a-9a^{2}=0$$

Tauwehea te a.

$$a\left(-45-9a\right)=0$$

Hei kimi otinga whārite, me whakaoti te a=0 me te -45-9a=0.

$$a=0 a=-5$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 9a^{2} mai i ngā taha e rua.

$$a-9a^{2}=46a$$

Tangohia te 46a mai i ngā taha e rua.

$$a-9a^{2}-46a=0$$

Pahekotia te a me -46a, ka -45a.

$$-45a-9a^{2}=0$$

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.

$$-9a^{2}-45a=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -9 mō a, -45 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-45\right)±\sqrt{\left(-45\right)^{2}}}{2\left(-9\right)}$$

Tuhia te pūtakerua o te \left(-45\right)^{2}.

$$a=\frac{-\left(-45\right)±45}{2\left(-9\right)}$$

Ko te tauaro o -45 ko 45.

$$a=\frac{45±45}{2\left(-9\right)}$$

Whakareatia 2 ki te -9.

$$a=\frac{45±45}{-18}$$

Nā, me whakaoti te whārite a=\frac{45±45}{-18} ina he tāpiri te ±. Tāpiri 45 ki te 45.

$$a=\frac{90}{-18}$$

Whakawehe 90 ki te -18.

$$a=-5$$

Nā, me whakaoti te whārite a=\frac{45±45}{-18} ina he tango te ±. Tango 45 mai i 45.

$$a=\frac{0}{-18}$$

Whakawehe 0 ki te -18.

$$a=0$$

Kua oti te whārite te whakatau.

$$a=-5 a=0$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 9a^{2} mai i ngā taha e rua.

$$a-9a^{2}=46a$$

Tangohia te 46a mai i ngā taha e rua.

$$a-9a^{2}-46a=0$$

Pahekotia te a me -46a, ka -45a.

$$-45a-9a^{2}=0$$

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.

$$-9a^{2}-45a=0$$

Whakawehea ngā taha e rua ki te -9.

$$\frac{-9a^{2}-45a}{-9}=\frac{0}{-9}$$

Mā te whakawehe ki te -9 ka wetekia te whakareanga ki te -9.

$$a^{2}+\left(-\frac{45}{-9}\right)a=\frac{0}{-9}$$

Whakawehe -45 ki te -9.

$$a^{2}+5a=\frac{0}{-9}$$

Whakawehe 0 ki te -9.

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

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.

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

Pūruatia \frac{5}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$a^{2}+5a+\frac{25}{4}=\frac{25}{4}$$

Tauwehea a^{2}+5a+\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}.

$$\left(a+\frac{5}{2}\right)^{2}=\frac{25}{4}$$

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

$$\sqrt{\left(a+\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Whakarūnātia.

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

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

$$a=0 a=-5$$