Whakaoti i te x^4+1/2x^2=17/8

$$\frac { x ^ { 4 } + 1 } { 2 x ^ { 2 } } = \frac { 17 } { 8 }$$

Whakaoti mō x

$$x=-\frac{1}{2}=-0.5$$

$$x=\frac{1}{2}=0.5$$

$$x=2$$

$$x=-2$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 8x^{2}, arā, te tauraro pātahi he tino iti rawa te kitea o 2x^{2},8.

$$4\left(x^{4}+1\right)=17x^{2}$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te x^{4}+1.

$$4x^{4}+4=17x^{2}$$

Tangohia te 17x^{2} mai i ngā taha e rua.

$$4x^{4}+4-17x^{2}=0$$

Whakakapia te t mō te x^{2}.

$$4t^{2}-17t+4=0$$

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 4 mō te a, te -17 mō te b, me te 4 mō te c i te ture pūrua.

$$t=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 4\times 4}}{2\times 4}$$

Mahia ngā tātaitai.

$$t=\frac{17±15}{8}$$

Whakaotia te whārite t=\frac{17±15}{8} ina he tōrunga te ±, ina he tōraro te ±.

$$t=4 t=\frac{1}{4}$$

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

$$x=2 x=-2 x=\frac{1}{2} x=-\frac{1}{2}$$