Whakaoti i te 6x^2-12x+3geq0

$$6 x ^ { 2 } - 12 x + 3 \geq 0$$

Whakaoti mō x

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

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.

$$6x^{2}-12x+3=0$$

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 6 mō te a, te -12 mō te b, me te 3 mō te c i te ture pūrua.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 6\times 3}}{2\times 6}$$

Mahia ngā tātaitai.

$$x=\frac{12±6\sqrt{2}}{12}$$

Whakaotia te whārite x=\frac{12±6\sqrt{2}}{12} ina he tōrunga te ±, ina he tōraro te ±.

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

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

$$6\left(x-\left(\frac{\sqrt{2}}{2}+1\right)\right)\left(x-\left(-\frac{\sqrt{2}}{2}+1\right)\right)\geq 0$$

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

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

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

$$x\leq -\frac{\sqrt{2}}{2}+1$$

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

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

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

$$x\geq \frac{\sqrt{2}}{2}+1$$

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

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