Whakaoti i te k^2-k-4>0

$$k ^ { 2 } - k - 4 > 0$$

Whakaoti mō k

$$k\in \left(-\infty,\frac{1-\sqrt{17}}{2}\right)\cup \left(\frac{\sqrt{17}+1}{2},\infty\right)$$

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.

$$k^{2}-k-4=0$$

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 1 mō te a, te -1 mō te b, me te -4 mō te c i te ture pūrua.

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

Mahia ngā tātaitai.

$$k=\frac{1±\sqrt{17}}{2}$$

Whakaotia te whārite k=\frac{1±\sqrt{17}}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$k=\frac{\sqrt{17}+1}{2} k=\frac{1-\sqrt{17}}{2}$$

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

$$\left(k-\frac{\sqrt{17}+1}{2}\right)\left(k-\frac{1-\sqrt{17}}{2}\right)>0$$

Kia tōrunga te otinga, me tōraro tahi te k-\frac{\sqrt{17}+1}{2} me te k-\frac{1-\sqrt{17}}{2}, me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te k-\frac{\sqrt{17}+1}{2} me te k-\frac{1-\sqrt{17}}{2}.

$$k-\frac{\sqrt{17}+1}{2}<0 k-\frac{1-\sqrt{17}}{2}<0$$

Te otinga e whakaea i ngā koreōrite e rua ko k<\frac{1-\sqrt{17}}{2}.

$$k<\frac{1-\sqrt{17}}{2}$$

Whakaarohia te tauira ina he tōrunga tahi te k-\frac{\sqrt{17}+1}{2} me te k-\frac{1-\sqrt{17}}{2}.

$$k-\frac{1-\sqrt{17}}{2}>0 k-\frac{\sqrt{17}+1}{2}>0$$

Te otinga e whakaea i ngā koreōrite e rua ko k>\frac{\sqrt{17}+1}{2}.

$$k>\frac{\sqrt{17}+1}{2}$$

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

$$k<\frac{1-\sqrt{17}}{2}\text{; }k>\frac{\sqrt{17}+1}{2}$$