Whakaoti i te k^2-24geq0

$$k ^ { 2 } - 24 \geq 0$$

Whakaoti mō k

$$k\in (-\infty,-2\sqrt{6}]\cup [2\sqrt{6},\infty)$$

Ngā Upane Otinga

Me tāpiri te 24 ki ngā taha e rua.

$$k^{2}\geq 24$$

Tātaitia te pūtakerua o 24 kia tae ki 2\sqrt{6}. Tuhia anō te 24 hei \left(2\sqrt{6}\right)^{2}.

$$k^{2}\geq \left(2\sqrt{6}\right)^{2}$$

E mau ana te koreōrite mō |k|\geq 2\sqrt{6}.

$$|k|\geq 2\sqrt{6}$$

Tuhia anō te |k|\geq 2\sqrt{6} hei k\leq -2\sqrt{6}\text{; }k\geq 2\sqrt{6}.

$$k\leq -2\sqrt{6}\text{; }k\geq 2\sqrt{6}$$