Whakaoti i te {l}{3x^2+2x}{-1geq0}

$$\left. \begin{array} { l } { 3 x ^ { 2 } + 2 x } \\ { - 1 \geq 0 } \end{array} \right.$$

Whakaoti mō x

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

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.

$$3x^{2}+2x-1=0$$

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 3 mō te a, te 2 mō te b, me te -1 mō te c i te ture pūrua.

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

Mahia ngā tātaitai.

$$x=\frac{-2±4}{6}$$

Whakaotia te whārite x=\frac{-2±4}{6} ina he tōrunga te ±, ina he tōraro te ±.

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

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

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

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

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

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

$$x\leq -1$$

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

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

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

$$x\geq \frac{1}{3}$$

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

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