Whakaoti i te x^4-16x^2-225=0

$$x ^ { 4 } - 16 x ^ { 2 } - 225 = 0$$

Whakaoti mō x (complex solution)

$$x=5$$

$$x=-5$$

$$x=-3i$$

$$x=3i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakakapia te t mō te x^{2}.

$$t^{2}-16t-225=0$$

Ka taea ngā whārite katoa o te momo ax^{2}+bx+c=0 te whakaoti mā te ture pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Whakakapia te 1 mō te a, te -16 mō te b, me te -225 mō te c i te ture pūrua.

$$t=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 1\left(-225\right)}}{2}$$

Mahia ngā tātaitai.

$$t=\frac{16±34}{2}$$

Whakaotia te whārite t=\frac{16±34}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$t=25 t=-9$$

I te mea ko x=t^{2}, ka riro ngā otinga mā te arotake i te x=±\sqrt{t} mō ia t.

$$x=-5 x=5 x=-3i x=3i$$

Whakaoti mō x

$$x=-5$$

$$x=5$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakakapia te t mō te x^{2}.

$$t^{2}-16t-225=0$$

Ka taea ngā whārite katoa o te momo ax^{2}+bx+c=0 te whakaoti mā te ture pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Whakakapia te 1 mō te a, te -16 mō te b, me te -225 mō te c i te ture pūrua.

$$t=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 1\left(-225\right)}}{2}$$

Mahia ngā tātaitai.

$$t=\frac{16±34}{2}$$

Whakaotia te whārite t=\frac{16±34}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$t=25 t=-9$$

I te mea ko x=t^{2}, ka riro ngā otinga mā te arotake i te x=±\sqrt{t} mō t tōrunga.

$$x=5 x=-5$$