Whakaoti i te 10x^2-7x-12geq0

$$10 x ^ { 2 } - 7 x - 12 \geq 0$$

Whakaoti mō x

$$x\in (-\infty,-\frac{4}{5}]\cup [\frac{3}{2},\infty)$$

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.

$$10x^{2}-7x-12=0$$

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 10 mō te a, te -7 mō te b, me te -12 mō te c i te ture pūrua.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 10\left(-12\right)}}{2\times 10}$$

Mahia ngā tātaitai.

$$x=\frac{7±23}{20}$$

Whakaotia te whārite x=\frac{7±23}{20} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{3}{2} x=-\frac{4}{5}$$

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

$$10\left(x-\frac{3}{2}\right)\left(x+\frac{4}{5}\right)\geq 0$$

Kia ≥0 te otinga, me ≤0 tahi, me ≥0 tahi rānei te x-\frac{3}{2} me te x+\frac{4}{5}. Whakaarohia te tauira ina he ≤0 tahi te x-\frac{3}{2} me te x+\frac{4}{5}.

$$x-\frac{3}{2}\leq 0 x+\frac{4}{5}\leq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\leq -\frac{4}{5}.

$$x\leq -\frac{4}{5}$$

Whakaarohia te tauira ina he ≥0 tahi te x-\frac{3}{2} me te x+\frac{4}{5}.

$$x+\frac{4}{5}\geq 0 x-\frac{3}{2}\geq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\geq \frac{3}{2}.

$$x\geq \frac{3}{2}$$

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

$$x\leq -\frac{4}{5}\text{; }x\geq \frac{3}{2}$$