Whakaoti i te x^2-3x-40geq0

$$x ^ { 2 } - 3 x - 40 \geq 0$$

Whakaoti mō x

$$x\in (-\infty,-5]\cup [8,\infty)$$

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.

$$x^{2}-3x-40=0$$

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 -3 mō te b, me te -40 mō te c i te ture pūrua.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\left(-40\right)}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{3±13}{2}$$

Whakaotia te whārite x=\frac{3±13}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$x=8 x=-5$$

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

$$\left(x-8\right)\left(x+5\right)\geq 0$$

Kia ≥0 te otinga, me ≤0 tahi, me ≥0 tahi rānei te x-8 me te x+5. Whakaarohia te tauira ina he ≤0 tahi te x-8 me te x+5.

$$x-8\leq 0 x+5\leq 0$$

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-8 me te x+5.

$$x+5\geq 0 x-8\geq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\geq 8.

$$x\geq 8$$

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

$$x\leq -5\text{; }x\geq 8$$