Whakaoti i te 4x(x-4)+7geq0

$$4 x ( x - 4 ) + 7 \geq 0$$

Whakaoti mō x

$$x\in (-\infty,\frac{1}{2}]\cup [\frac{7}{2},\infty)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 4x ki te x-4.

$$4x^{2}-16x+7\geq 0$$

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.

$$4x^{2}-16x+7=0$$

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 -16 mō te b, me te 7 mō te c i te ture pūrua.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 4\times 7}}{2\times 4}$$

Mahia ngā tātaitai.

$$x=\frac{16±12}{8}$$

Whakaotia te whārite x=\frac{16±12}{8} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{7}{2} x=\frac{1}{2}$$

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

$$4\left(x-\frac{7}{2}\right)\left(x-\frac{1}{2}\right)\geq 0$$

Kia ≥0 te otinga, me ≤0 tahi, me ≥0 tahi rānei te x-\frac{7}{2} me te x-\frac{1}{2}. Whakaarohia te tauira ina he ≤0 tahi te x-\frac{7}{2} me te x-\frac{1}{2}.

$$x-\frac{7}{2}\leq 0 x-\frac{1}{2}\leq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\leq \frac{1}{2}.

$$x\leq \frac{1}{2}$$

Whakaarohia te tauira ina he ≥0 tahi te x-\frac{7}{2} me te x-\frac{1}{2}.

$$x-\frac{1}{2}\geq 0 x-\frac{7}{2}\geq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\geq \frac{7}{2}.

$$x\geq \frac{7}{2}$$

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

$$x\leq \frac{1}{2}\text{; }x\geq \frac{7}{2}$$