Whakaoti i te 2x^2-5x+2geq0

$$2 { x }^{ 2 } -5x+2 \geq 0$$

Whakaoti mō x

$$x\in (-\infty,\frac{1}{2}]\cup [2,\infty)$$

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.

$$2x^{2}-5x+2=0$$

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

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

Mahia ngā tātaitai.

$$x=\frac{5±3}{4}$$

Whakaotia te whārite x=\frac{5±3}{4} ina he tōrunga te ±, ina he tōraro te ±.

$$x=2 x=\frac{1}{2}$$

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

$$2\left(x-2\right)\left(x-\frac{1}{2}\right)\geq 0$$

Kia ≥0 te otinga, me ≤0 tahi, me ≥0 tahi rānei te x-2 me te x-\frac{1}{2}. Whakaarohia te tauira ina he ≤0 tahi te x-2 me te x-\frac{1}{2}.

$$x-2\leq 0 x-\frac{1}{2}\leq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\leq \frac{1}{2}.

$$x\leq \frac{1}{2}$$

Whakaarohia te tauira ina he ≥0 tahi te x-2 me te x-\frac{1}{2}.

$$x-\frac{1}{2}\geq 0 x-2\geq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\geq 2.

$$x\geq 2$$

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

$$x\leq \frac{1}{2}\text{; }x\geq 2$$