Whakaoti i te 5x^2+4xgeq0

$$5 x ^ { 2 } + 4 x \geq 0$$

Whakaoti mō x

$$x\in (-\infty,-\frac{4}{5}]\cup [0,\infty)$$

Ngā Upane Otinga

Tauwehea te x.

$$x\left(5x+4\right)\geq 0$$

Kia ≥0 te otinga, me ≤0 tahi, me ≥0 tahi rānei te x+\frac{4}{5} me te x. Whakaarohia te tauira ina he ≤0 tahi te x+\frac{4}{5} me te x.

$$x+\frac{4}{5}\leq 0 x\leq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\leq -\frac{4}{5}.

$$x\leq -\frac{4}{5}$$

Whakaarohia te tauira ina he ≥0 tahi te x+\frac{4}{5} me te x.

$$x\geq 0 x+\frac{4}{5}\geq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\geq 0.

$$x\geq 0$$

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

$$x\leq -\frac{4}{5}\text{; }x\geq 0$$