Question-158410

Question Number 158410 by tebohlouis last updated on 03/Nov/21

Answered by MJS_new last updated on 03/Nov/21

2 x + 3 ⩾ 0 \land (x - 2) (x + 1) > 0 \lor 2 x + 3 ⩽ 0 \land (x - 2) (x + 1) < 0

2 x + 3 ⩾ 0 \Rightarrow x ⩾ - \frac{3}{2}

(x - 2) (x + 1) > 0 \Rightarrow x < - 1 \lor x > 2

\Rightarrow - \frac{3}{2} ⩽ x < - 1 \lor x > 2

2 x + 3 ⩽ 0 \Rightarrow x ⩽ - \frac{3}{2}

(x - 2) (x + 1) < 0 \Rightarrow - 1 < x < 2

no solution

\Rightarrow x \in [- \frac{3}{2}; - 1) \cup (2; + \infty)

Answered by physicstutes last updated on 06/Nov/21

{zero}^{'} s

x - 2 = 0 \Rightarrow x = 2

x + 1 = 0 \Rightarrow x = - 1

2 x + 3 = 0 \Rightarrow x = - \frac{3}{2}

\begin{array}{cc} x < - \frac{3}{2} & - \frac{3}{2} < x < - 1 & - 1 < x < 2 & x > 2 \\ - & + & - & + \end{array}

S = {x : - \frac{3}{2} ⩽ x < - 1 \cup x > 2}