Whakaoti i te 23x^2+12x-35geq0

$$23 x ^ { 2 } + 12 x - 35 \geq 0$$

Whakaoti mō x

$$x\in (-\infty,-\frac{35}{23}]\cup [1,\infty)$$

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.

$$23x^{2}+12x-35=0$$

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 23 mō te a, te 12 mō te b, me te -35 mō te c i te ture pūrua.

$$x=\frac{-12±\sqrt{12^{2}-4\times 23\left(-35\right)}}{2\times 23}$$

Mahia ngā tātaitai.

$$x=\frac{-12±58}{46}$$

Whakaotia te whārite x=\frac{-12±58}{46} ina he tōrunga te ±, ina he tōraro te ±.

$$x=1 x=-\frac{35}{23}$$

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

$$23\left(x-1\right)\left(x+\frac{35}{23}\right)\geq 0$$

Kia ≥0 te otinga, me ≤0 tahi, me ≥0 tahi rānei te x-1 me te x+\frac{35}{23}. Whakaarohia te tauira ina he ≤0 tahi te x-1 me te x+\frac{35}{23}.

$$x-1\leq 0 x+\frac{35}{23}\leq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\leq -\frac{35}{23}.

$$x\leq -\frac{35}{23}$$

Whakaarohia te tauira ina he ≥0 tahi te x-1 me te x+\frac{35}{23}.

$$x+\frac{35}{23}\geq 0 x-1\geq 0$$

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.

$$x\leq -\frac{35}{23}\text{; }x\geq 1$$