Whakaoti i te x^2+2x-15geq0

$$x ^ { 2 } + 2 x - 15 \geq 0$$

Whakaoti mō x

$$x\in (-\infty,-5]\cup [3,\infty)$$

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.

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

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 2 mō te b, me te -15 mō te c i te ture pūrua.

$$x=\frac{-2±\sqrt{2^{2}-4\times 1\left(-15\right)}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{-2±8}{2}$$

Whakaotia te whārite x=\frac{-2±8}{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.

$$\left(x-3\right)\left(x+5\right)\geq 0$$

Kia ≥0 te otinga, me ≤0 tahi, me ≥0 tahi rānei te x-3 me te x+5. Whakaarohia te tauira ina he ≤0 tahi te x-3 me te x+5.

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

Te otinga e whakaea i ngā koreōrite e rua ko x\leq -5.

$$x\leq -5$$

Whakaarohia te tauira ina he ≥0 tahi te x-3 me te x+5.

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

Te otinga e whakaea i ngā koreōrite e rua ko x\geq 3.

$$x\geq 3$$

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

$$x\leq -5\text{; }x\geq 3$$