Whakaoti i te x^2+8x+15leq0

$$x ^ { 2 } + 8 x + 15 \leq 0$$

Whakaoti mō x

$$x\in \begin{bmatrix}-5,-3\end{bmatrix}$$

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.

$$x^{2}+8x+15=0$$

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 8 mō te b, me te 15 mō te c i te ture pūrua.

$$x=\frac{-8±\sqrt{8^{2}-4\times 1\times 15}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{-8±2}{2}$$

Whakaotia te whārite x=\frac{-8±2}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$x=-3 x=-5$$

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

$$\left(x+3\right)\left(x+5\right)\leq 0$$

Kia ≤0 te otinga, me ≥0 rawa tētahi uara o x+3 me x+5, me ≤0 anō te uara o tētahi. Whakaarohia te tauira ina ko x+3\geq 0 me x+5\leq 0.

$$x+3\geq 0 x+5\leq 0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina ko x+3\leq 0 me x+5\geq 0.

$$x+5\geq 0 x+3\leq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left[-5,-3\right].

$$x\in \begin{bmatrix}-5,-3\end{bmatrix}$$

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

$$x\in \begin{bmatrix}-5,-3\end{bmatrix}$$