2x+8<x^2

Solution

2 x + 8 < x^{2}

x < - 2 or x > 4

Interval Notation

(- \infty, - 2) \cup (4, \infty)

Solution steps

2 x + 8 < x^{2}

Rewrite in standard form

2 x + 8 - x^{2} < 0

Factor

2 x + 8 - x^{2} : - (x + 2) (x - 4)

- (x + 2) (x - 4) < 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x + 2) (x - 4)) (- 1) > 0 \cdot (- 1)

Simplify

(x + 2) (x - 4) > 0

Identify the intervals

x < - 2 or x > 4

240 = w (2 w - 4)

\frac{1}{4} n^{2} + 3 = 4

7^{2} + b^{2} = 2 5^{2}

4 x + 2 + x^{2} = 0

2 x^{3} - x^{4} \leq 0