k^2-24geq0 eching

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

k uchun yechish

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

Yechim qadamlari

24 ni ikki tarafga qo’shing.

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

24 ning kvadrat ildizini hisoblab, 2\sqrt{6} natijaga ega bo‘ling. 24 ni \left(2\sqrt{6}\right)^{2} sifatida qaytadan yozish.

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

Tengsizlikda |k|\geq 2\sqrt{6} bor.

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

|k|\geq 2\sqrt{6} ni k\leq -2\sqrt{6}\text{; }k\geq 2\sqrt{6} sifatida qaytadan yozish.

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