Question 444857


{{{x^2 - 5x ≥ -6 }}}....factor

{{{x^2 - 5x+6 ≥ 0 }}}

{{{x^2 - 2x-3x +6 ≥ 0 }}}

{{{(x^2 - 2x)-(3x -6) ≥ 0 }}}

{{{x(x - 2)-3(x -2) ≥ 0 }}}

{{{(x-3)(x -2) ≥ 0 }}}

solutions:

{{{(x-3) ≥ 0 }}}...=>{{{x ≥ 3}}}

or

{{{(x-2) <= 0 }}}...=>{{{x <= 2}}}


so, the answer is: B. (-&#8734;, 2] or [3, &#8734;)