Whakaoti i te 6x^2-13x-5geq0

$$6 x ^ { 2 } - 13 x - 5 \geq 0$$

Whakaoti mō x

$$x\in (-\infty,-\frac{1}{3}]\cup [\frac{5}{2},\infty)$$

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.

$$6x^{2}-13x-5=0$$

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 6 mō te a, te -13 mō te b, me te -5 mō te c i te ture pūrua.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 6\left(-5\right)}}{2\times 6}$$

Mahia ngā tātaitai.

$$x=\frac{13±17}{12}$$

Whakaotia te whārite x=\frac{13±17}{12} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{5}{2} x=-\frac{1}{3}$$

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

$$6\left(x-\frac{5}{2}\right)\left(x+\frac{1}{3}\right)\geq 0$$

Kia ≥0 te otinga, me ≤0 tahi, me ≥0 tahi rānei te x-\frac{5}{2} me te x+\frac{1}{3}. Whakaarohia te tauira ina he ≤0 tahi te x-\frac{5}{2} me te x+\frac{1}{3}.

$$x-\frac{5}{2}\leq 0 x+\frac{1}{3}\leq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\leq -\frac{1}{3}.

$$x\leq -\frac{1}{3}$$

Whakaarohia te tauira ina he ≥0 tahi te x-\frac{5}{2} me te x+\frac{1}{3}.

$$x+\frac{1}{3}\geq 0 x-\frac{5}{2}\geq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\geq \frac{5}{2}.

$$x\geq \frac{5}{2}$$

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

$$x\leq -\frac{1}{3}\text{; }x\geq \frac{5}{2}$$