Whakaoti i te 5a^2-5a-2geq0

$$5 a ^ { 2 } - 5 a - 2 \geq 0$$

Whakaoti mō a

$$a\in (-\infty,-\frac{\sqrt{65}}{10}+\frac{1}{2}]\cup [\frac{\sqrt{65}}{10}+\frac{1}{2},\infty)$$

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.

$$5a^{2}-5a-2=0$$

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 5 mō te a, te -5 mō te b, me te -2 mō te c i te ture pūrua.

$$a=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 5\left(-2\right)}}{2\times 5}$$

Mahia ngā tātaitai.

$$a=\frac{5±\sqrt{65}}{10}$$

Whakaotia te whārite a=\frac{5±\sqrt{65}}{10} ina he tōrunga te ±, ina he tōraro te ±.

$$a=\frac{\sqrt{65}}{10}+\frac{1}{2} a=-\frac{\sqrt{65}}{10}+\frac{1}{2}$$

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

$$5\left(a-\left(\frac{\sqrt{65}}{10}+\frac{1}{2}\right)\right)\left(a-\left(-\frac{\sqrt{65}}{10}+\frac{1}{2}\right)\right)\geq 0$$

Kia ≥0 te otinga, me ≤0 tahi, me ≥0 tahi rānei te a-\left(\frac{\sqrt{65}}{10}+\frac{1}{2}\right) me te a-\left(-\frac{\sqrt{65}}{10}+\frac{1}{2}\right). Whakaarohia te tauira ina he ≤0 tahi te a-\left(\frac{\sqrt{65}}{10}+\frac{1}{2}\right) me te a-\left(-\frac{\sqrt{65}}{10}+\frac{1}{2}\right).

$$a-\left(\frac{\sqrt{65}}{10}+\frac{1}{2}\right)\leq 0 a-\left(-\frac{\sqrt{65}}{10}+\frac{1}{2}\right)\leq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko a\leq -\frac{\sqrt{65}}{10}+\frac{1}{2}.

$$a\leq -\frac{\sqrt{65}}{10}+\frac{1}{2}$$

Whakaarohia te tauira ina he ≥0 tahi te a-\left(\frac{\sqrt{65}}{10}+\frac{1}{2}\right) me te a-\left(-\frac{\sqrt{65}}{10}+\frac{1}{2}\right).

$$a-\left(-\frac{\sqrt{65}}{10}+\frac{1}{2}\right)\geq 0 a-\left(\frac{\sqrt{65}}{10}+\frac{1}{2}\right)\geq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko a\geq \frac{\sqrt{65}}{10}+\frac{1}{2}.

$$a\geq \frac{\sqrt{65}}{10}+\frac{1}{2}$$

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

$$a\leq -\frac{\sqrt{65}}{10}+\frac{1}{2}\text{; }a\geq \frac{\sqrt{65}}{10}+\frac{1}{2}$$