Question

Решите неравенства (12+x)^2 ≥ x^2+21​

Answer 1

$(12+x)^2\geq x^2+21\\144+24x+x^2-x^2\geq 21\\24x\geq 21-144\\24x\geq -123\ /:24\\x\geq -\frac{41}8}\\x\in[-\frac{41}{8};+\infty)$