Question

Solve the equation: $x^2-5x+6\ge 0$.

• A. $x\in (-\infty ,2]\cup [3,\infty )$
• B. $x\in [2,3]$
• C. $x\in (-\infty ,-1]\cup [6,\infty )$
• D. $x\in [-1,6]$

Answer 1

The correct option is A $x\in (-\infty ,2]\cup [3,\infty )$

$x^2-5x+6\ge 0$
$\Rightarrow x^2-3x-2x+6\ge 0$
$\Rightarrow x(x-3)-2(x-3)\ge 0$
$\Rightarrow (x-3)(x-2)\ge 0$
There are two possibilities
$(x-3)\ge 0,(x-2)\ge 0$
or $(x-3)\le 0,(x-2)\le 0$
$\therefore x\in (-\infty ,2]\cup [3,\infty )$