Datrys k^2-13geq0

$$k ^ { 2 } - 13 \geq 0$$

Datrys ar gyfer k

$$k\in (-\infty,-\sqrt{13}]\cup [\sqrt{13},\infty)$$

Camau Ateb

Ychwanegu 13 at y ddwy ochr.

$$k^{2}\geq 13$$

Cyfrifo ail isradd 13 a chael \sqrt{13}. Ailysgrifennwch 13 fel \left(\sqrt{13}\right)^{2}.

$$k^{2}\geq \left(\sqrt{13}\right)^{2}$$

Mae anghydraddoldeb ar gyfer |k|\geq \sqrt{13}.

$$|k|\geq \sqrt{13}$$

Ailysgrifennwch |k|\geq \sqrt{13} fel k\leq -\sqrt{13}\text{; }k\geq \sqrt{13}.

$$k\leq -\sqrt{13}\text{; }k\geq \sqrt{13}$$