Whakaoti i te 2x^2-13x+11leq0

$$2 x ^ { 2 } - 13 x + 11 \leq 0$$

Whakaoti mō x

$$x\in \begin{bmatrix}1,\frac{11}{2}\end{bmatrix}$$

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.

$$2x^{2}-13x+11=0$$

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 -13 mō te b, me te 11 mō te c i te ture pūrua.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 2\times 11}}{2\times 2}$$

Mahia ngā tātaitai.

$$x=\frac{13±9}{4}$$

Whakaotia te whārite x=\frac{13±9}{4} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{11}{2} x=1$$

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

$$2\left(x-\frac{11}{2}\right)\left(x-1\right)\leq 0$$

Kia ≤0 te otinga, me ≥0 rawa tētahi uara o x-\frac{11}{2} me x-1, me ≤0 anō te uara o tētahi. Whakaarohia te tauira ina ko x-\frac{11}{2}\geq 0 me x-1\leq 0.

$$x-\frac{11}{2}\geq 0 x-1\leq 0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina ko x-\frac{11}{2}\leq 0 me x-1\geq 0.

$$x-1\geq 0 x-\frac{11}{2}\leq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left[1,\frac{11}{2}\right].

$$x\in \begin{bmatrix}1,\frac{11}{2}\end{bmatrix}$$

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

$$x\in \begin{bmatrix}1,\frac{11}{2}\end{bmatrix}$$