Whakaoti i te x^2-x-16<0

$$x ^ { 2 } - x - 16 < 0$$

Whakaoti mō x

$$x\in \left(\frac{1-\sqrt{65}}{2},\frac{\sqrt{65}+1}{2}\right)$$

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.

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

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 -1 mō te b, me te -16 mō te c i te ture pūrua.

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

Mahia ngā tātaitai.

$$x=\frac{1±\sqrt{65}}{2}$$

Whakaotia te whārite x=\frac{1±\sqrt{65}}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{\sqrt{65}+1}{2} x=\frac{1-\sqrt{65}}{2}$$

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

$$\left(x-\frac{\sqrt{65}+1}{2}\right)\left(x-\frac{1-\sqrt{65}}{2}\right)<0$$

Kia tōraro te otinga, me tauaro rawa ngā tohu o te x-\frac{\sqrt{65}+1}{2} me te x-\frac{1-\sqrt{65}}{2}. Whakaarohia te tauira ina he tōrunga te x-\frac{\sqrt{65}+1}{2} he tōraro te x-\frac{1-\sqrt{65}}{2}.

$$x-\frac{\sqrt{65}+1}{2}>0 x-\frac{1-\sqrt{65}}{2}<0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina he tōrunga te x-\frac{1-\sqrt{65}}{2} he tōraro te x-\frac{\sqrt{65}+1}{2}.

$$x-\frac{1-\sqrt{65}}{2}>0 x-\frac{\sqrt{65}+1}{2}<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left(\frac{1-\sqrt{65}}{2},\frac{\sqrt{65}+1}{2}\right).

$$x\in \left(\frac{1-\sqrt{65}}{2},\frac{\sqrt{65}+1}{2}\right)$$

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

$$x\in \left(\frac{1-\sqrt{65}}{2},\frac{\sqrt{65}+1}{2}\right)$$