Whakaoti i te 2x^2+4x-2leq0

$$2 x ^ { 2 } + 4 x - 2 \leq 0$$

Whakaoti mō x

$$x\in \begin{bmatrix}-\sqrt{2}-1,\sqrt{2}-1\end{bmatrix}$$

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.

$$2x^{2}+4x-2=0$$

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

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

Mahia ngā tātaitai.

$$x=\frac{-4±4\sqrt{2}}{4}$$

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

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

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

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

Kia ≤0 te otinga, me ≥0 rawa tētahi uara o x-\left(\sqrt{2}-1\right) me x-\left(-\sqrt{2}-1\right), me ≤0 anō te uara o tētahi. Whakaarohia te tauira ina ko x-\left(\sqrt{2}-1\right)\geq 0 me x-\left(-\sqrt{2}-1\right)\leq 0.

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

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina ko x-\left(\sqrt{2}-1\right)\leq 0 me x-\left(-\sqrt{2}-1\right)\geq 0.

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

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left[-\left(\sqrt{2}+1\right),\sqrt{2}-1\right].

$$x\in \begin{bmatrix}-\left(\sqrt{2}+1\right),\sqrt{2}-1\end{bmatrix}$$

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

$$x\in \begin{bmatrix}-\sqrt{2}-1,\sqrt{2}-1\end{bmatrix}$$