Whakaoti i te t^2-6t+1geq0

$$t ^ { 2 } - 6 t + 1 \geq 0$$

Whakaoti mō t

$$t\in (-\infty,3-2\sqrt{2}]\cup [2\sqrt{2}+3,\infty)$$

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.

$$t^{2}-6t+1=0$$

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 -6 mō te b, me te 1 mō te c i te ture pūrua.

$$t=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\times 1}}{2}$$

Mahia ngā tātaitai.

$$t=\frac{6±4\sqrt{2}}{2}$$

Whakaotia te whārite t=\frac{6±4\sqrt{2}}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$t=2\sqrt{2}+3 t=3-2\sqrt{2}$$

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

$$\left(t-\left(2\sqrt{2}+3\right)\right)\left(t-\left(3-2\sqrt{2}\right)\right)\geq 0$$

Kia ≥0 te otinga, me ≤0 tahi, me ≥0 tahi rānei te t-\left(2\sqrt{2}+3\right) me te t-\left(3-2\sqrt{2}\right). Whakaarohia te tauira ina he ≤0 tahi te t-\left(2\sqrt{2}+3\right) me te t-\left(3-2\sqrt{2}\right).

$$t-\left(2\sqrt{2}+3\right)\leq 0 t-\left(3-2\sqrt{2}\right)\leq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko t\leq 3-2\sqrt{2}.

$$t\leq 3-2\sqrt{2}$$

Whakaarohia te tauira ina he ≥0 tahi te t-\left(2\sqrt{2}+3\right) me te t-\left(3-2\sqrt{2}\right).

$$t-\left(3-2\sqrt{2}\right)\geq 0 t-\left(2\sqrt{2}+3\right)\geq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko t\geq 2\sqrt{2}+3.

$$t\geq 2\sqrt{2}+3$$

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

$$t\leq 3-2\sqrt{2}\text{; }t\geq 2\sqrt{2}+3$$