Whakaoti i te x^2-3x+1<0

$${ x }^{ 2 } -3x+1 < 0$$

Whakaoti mō x

$$x\in \left(\frac{3-\sqrt{5}}{2},\frac{\sqrt{5}+3}{2}\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Kia whakaotia te koreōrite, me tauwehe te taha mauī. Ka taea te huamaha pūrua te tauwehe mā te whakamahi i te huringa ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), ina ko x_{1} me x_{2} ngā otinga o te whārite pūrua ax^{2}+bx+c=0.

$$x^{2}-3x+1=0$$

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 -3 mō te b, me te 1 mō te c i te ture pūrua.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 1}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{3±\sqrt{5}}{2}$$

Whakaotia te whārite x=\frac{3±\sqrt{5}}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{\sqrt{5}+3}{2} x=\frac{3-\sqrt{5}}{2}$$

Tuhia anō te koreōrite mā te whakamahi i ngā otinga i whiwhi.

$$\left(x-\frac{\sqrt{5}+3}{2}\right)\left(x-\frac{3-\sqrt{5}}{2}\right)<0$$

Kia tōraro te otinga, me tauaro rawa ngā tohu o te x-\frac{\sqrt{5}+3}{2} me te x-\frac{3-\sqrt{5}}{2}. Whakaarohia te tauira ina he tōrunga te x-\frac{\sqrt{5}+3}{2} he tōraro te x-\frac{3-\sqrt{5}}{2}.

$$x-\frac{\sqrt{5}+3}{2}>0 x-\frac{3-\sqrt{5}}{2}<0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina he tōrunga te x-\frac{3-\sqrt{5}}{2} he tōraro te x-\frac{\sqrt{5}+3}{2}.

$$x-\frac{3-\sqrt{5}}{2}>0 x-\frac{\sqrt{5}+3}{2}<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left(\frac{3-\sqrt{5}}{2},\frac{\sqrt{5}+3}{2}\right).

$$x\in \left(\frac{3-\sqrt{5}}{2},\frac{\sqrt{5}+3}{2}\right)$$

Ko te otinga whakamutunga ko te whakakotahi i ngā otinga kua whiwhi.

$$x\in \left(\frac{3-\sqrt{5}}{2},\frac{\sqrt{5}+3}{2}\right)$$