Whakaoti i te 2x^2+10x-12geq0

$$2 x ^ { 2 } + 10 x - 12 \geq 0$$

Whakaoti mō x

$$x\in (-\infty,-6]\cup [1,\infty)$$

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.

$$2x^{2}+10x-12=0$$

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 10 mō te b, me te -12 mō te c i te ture pūrua.

$$x=\frac{-10±\sqrt{10^{2}-4\times 2\left(-12\right)}}{2\times 2}$$

Mahia ngā tātaitai.

$$x=\frac{-10±14}{4}$$

Whakaotia te whārite x=\frac{-10±14}{4} ina he tōrunga te ±, ina he tōraro te ±.

$$x=1 x=-6$$

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

$$2\left(x-1\right)\left(x+6\right)\geq 0$$

Kia ≥0 te otinga, me ≤0 tahi, me ≥0 tahi rānei te x-1 me te x+6. Whakaarohia te tauira ina he ≤0 tahi te x-1 me te x+6.

$$x-1\leq 0 x+6\leq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\leq -6.

$$x\leq -6$$

Whakaarohia te tauira ina he ≥0 tahi te x-1 me te x+6.

$$x+6\geq 0 x-1\geq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\geq 1.

$$x\geq 1$$

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

$$x\leq -6\text{; }x\geq 1$$