x^2+x>= 42

Solution

x^{2} + x \geq 42

x \leq - 7 or x \geq 6

Interval Notation

(- \infty, - 7] \cup [6, \infty)

Solution steps

x^{2} + x \geq 42

Rewrite in standard form

x^{2} + x - 42 \geq 0

Factor

x^{2} + x - 42 : (x - 6) (x + 7)

(x - 6) (x + 7) \geq 0

Identify the intervals

x \leq - 7 or x \geq 6

2 d + 4 = 10 + 2.5 d

28 \cdot 2

- x + \frac{7}{2} < \frac{- 6 x - 3}{4} \leq \frac{3}{2} x + \frac{4}{5}

lo g_{2} (\frac{1}{64})