Whakaoti i te (x-2)^2geq7

$$( x - 2 ) ^ { 2 } \geq 7$$

Whakaoti mō x

$$x\in (-\infty,2-\sqrt{7}]\cup [\sqrt{7}+2,\infty)$$

Ngā Upane Otinga

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.

$$\left(x-2\right)^{2}=0$$

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 1 mō te a, te -4 mō te b, me te -3 mō te c i te ture pūrua.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\left(-3\right)}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{4±2\sqrt{7}}{2}$$

Whakaotia te whārite x=\frac{4±2\sqrt{7}}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\sqrt{7}+2 x=2-\sqrt{7}$$

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

$$\left(x-\left(\sqrt{7}+2\right)\right)\left(x-\left(2-\sqrt{7}\right)\right)\geq 0$$

Kia ≥0 te otinga, me ≤0 tahi, me ≥0 tahi rānei te x-\left(\sqrt{7}+2\right) me te x-\left(2-\sqrt{7}\right). Whakaarohia te tauira ina he ≤0 tahi te x-\left(\sqrt{7}+2\right) me te x-\left(2-\sqrt{7}\right).

$$x-\left(\sqrt{7}+2\right)\leq 0 x-\left(2-\sqrt{7}\right)\leq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\leq 2-\sqrt{7}.

$$x\leq 2-\sqrt{7}$$

Whakaarohia te tauira ina he ≥0 tahi te x-\left(\sqrt{7}+2\right) me te x-\left(2-\sqrt{7}\right).

$$x-\left(2-\sqrt{7}\right)\geq 0 x-\left(\sqrt{7}+2\right)\geq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\geq \sqrt{7}+2.

$$x\geq \sqrt{7}+2$$

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

$$x\leq 2-\sqrt{7}\text{; }x\geq \sqrt{7}+2$$