Whakaoti i te x^2+11x-10geq0

$${ x }^{ 2 } +11x-10 \geq 0$$

Whakaoti mō x

$$x\in (-\infty,\frac{-\sqrt{161}-11}{2}]\cup [\frac{\sqrt{161}-11}{2},\infty)$$

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.

$$x^{2}+11x-10=0$$

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 1 mō te a, te 11 mō te b, me te -10 mō te c i te ture pūrua.

$$x=\frac{-11±\sqrt{11^{2}-4\times 1\left(-10\right)}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{-11±\sqrt{161}}{2}$$

Whakaotia te whārite x=\frac{-11±\sqrt{161}}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{\sqrt{161}-11}{2} x=\frac{-\sqrt{161}-11}{2}$$

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

$$\left(x-\frac{\sqrt{161}-11}{2}\right)\left(x-\frac{-\sqrt{161}-11}{2}\right)\geq 0$$

Kia ≥0 te otinga, me ≤0 tahi, me ≥0 tahi rānei te x-\frac{\sqrt{161}-11}{2} me te x-\frac{-\sqrt{161}-11}{2}. Whakaarohia te tauira ina he ≤0 tahi te x-\frac{\sqrt{161}-11}{2} me te x-\frac{-\sqrt{161}-11}{2}.

$$x-\frac{\sqrt{161}-11}{2}\leq 0 x-\frac{-\sqrt{161}-11}{2}\leq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\leq \frac{-\sqrt{161}-11}{2}.

$$x\leq \frac{-\sqrt{161}-11}{2}$$

Whakaarohia te tauira ina he ≥0 tahi te x-\frac{\sqrt{161}-11}{2} me te x-\frac{-\sqrt{161}-11}{2}.

$$x-\frac{-\sqrt{161}-11}{2}\geq 0 x-\frac{\sqrt{161}-11}{2}\geq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\geq \frac{\sqrt{161}-11}{2}.

$$x\geq \frac{\sqrt{161}-11}{2}$$

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

$$x\leq \frac{-\sqrt{161}-11}{2}\text{; }x\geq \frac{\sqrt{161}-11}{2}$$