Whakaoti i te (2x)^2-12(x+1)geq0

$$( 2 x ) ^ { 2 } - 12 ( x + 1 ) \geq 0$$

Whakaoti mō x

$$x\in (-\infty,\frac{3-\sqrt{21}}{2}]\cup [\frac{\sqrt{21}+3}{2},\infty)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakarohaina te \left(2x\right)^{2}.

$$2^{2}x^{2}-12\left(x+1\right)\geq 0$$

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

$$4x^{2}-12\left(x+1\right)\geq 0$$

Whakamahia te āhuatanga tohatoha hei whakarea te -12 ki te x+1.

$$4x^{2}-12x-12\geq 0$$

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.

$$4x^{2}-12x-12=0$$

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

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

Mahia ngā tātaitai.

$$x=\frac{12±4\sqrt{21}}{8}$$

Whakaotia te whārite x=\frac{12±4\sqrt{21}}{8} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{\sqrt{21}+3}{2} x=\frac{3-\sqrt{21}}{2}$$

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

$$4\left(x-\frac{\sqrt{21}+3}{2}\right)\left(x-\frac{3-\sqrt{21}}{2}\right)\geq 0$$

Kia ≥0 te otinga, me ≤0 tahi, me ≥0 tahi rānei te x-\frac{\sqrt{21}+3}{2} me te x-\frac{3-\sqrt{21}}{2}. Whakaarohia te tauira ina he ≤0 tahi te x-\frac{\sqrt{21}+3}{2} me te x-\frac{3-\sqrt{21}}{2}.

$$x-\frac{\sqrt{21}+3}{2}\leq 0 x-\frac{3-\sqrt{21}}{2}\leq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\leq \frac{3-\sqrt{21}}{2}.

$$x\leq \frac{3-\sqrt{21}}{2}$$

Whakaarohia te tauira ina he ≥0 tahi te x-\frac{\sqrt{21}+3}{2} me te x-\frac{3-\sqrt{21}}{2}.

$$x-\frac{3-\sqrt{21}}{2}\geq 0 x-\frac{\sqrt{21}+3}{2}\geq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\geq \frac{\sqrt{21}+3}{2}.

$$x\geq \frac{\sqrt{21}+3}{2}$$

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

$$x\leq \frac{3-\sqrt{21}}{2}\text{; }x\geq \frac{\sqrt{21}+3}{2}$$