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

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

Whakaoti mō x

$$x\in \begin{bmatrix}-\frac{2}{3},\frac{1}{2}\end{bmatrix}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Me whakarea te koreōrite ki te -1 kia tōrunga ai te tau whakarea o te pū tino teitei i -6x^{2}-x+2. I te mea he tōraro a -1, ka huri te ahunga koreōrite.

$$6x^{2}+x-2\leq 0$$

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.

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

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 6 mō te a, te 1 mō te b, me te -2 mō te c i te ture pūrua.

$$x=\frac{-1±\sqrt{1^{2}-4\times 6\left(-2\right)}}{2\times 6}$$

Mahia ngā tātaitai.

$$x=\frac{-1±7}{12}$$

Whakaotia te whārite x=\frac{-1±7}{12} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{1}{2} x=-\frac{2}{3}$$

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

$$6\left(x-\frac{1}{2}\right)\left(x+\frac{2}{3}\right)\leq 0$$

Kia ≤0 te otinga, me ≥0 rawa tētahi uara o x-\frac{1}{2} me x+\frac{2}{3}, me ≤0 anō te uara o tētahi. Whakaarohia te tauira ina ko x-\frac{1}{2}\geq 0 me x+\frac{2}{3}\leq 0.

$$x-\frac{1}{2}\geq 0 x+\frac{2}{3}\leq 0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina ko x-\frac{1}{2}\leq 0 me x+\frac{2}{3}\geq 0.

$$x+\frac{2}{3}\geq 0 x-\frac{1}{2}\leq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left[-\frac{2}{3},\frac{1}{2}\right].

$$x\in \begin{bmatrix}-\frac{2}{3},\frac{1}{2}\end{bmatrix}$$

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

$$x\in \begin{bmatrix}-\frac{2}{3},\frac{1}{2}\end{bmatrix}$$