Question 1094168
<br> Some inspection shows that your answer can't be right.<br>
If x is any negative number, then both terms on the left are negative, so their sum is negative; therefore the sum is always less than 6.  So there are no negative values that satisfy the inequality.  In particular, the [-3,0) part of your solution set is not right.<br>
The other part of your solution set excludes the positive values less than 3.  But for x=1 we get 1+9= 10, which is greater than 6.  So that part of your answer is not right, either.<br>
But just getting a right answer is of little use.  What you should be after is understanding how to get the right answer.  For us to help you, you need to show us, when you post your question, what you did to try to solve the problem.<br>
In order to solve this kind of inequality, you need to multiply both sides by x to get a quadratic inequality.  What most beginning students don't realize is that you have to work with two separate cases, because if x is negative, the direction of the inequality changes.<br>
With your example, we start with
{{{x + 9/x >= 6}}}<br>
When we multiply both sides by x to get a quadratic inequality, we need to break the process into two cases:
(1) IF x > 0 (we know it can't be EQUAL to 0), THEN
{{{x^2 + 9 >= 6x}}}
{{{x^2 - 6x + 9 >= 0}}}
{{{(x-3)^2 >= 0}}}<br>
But any expression, when squared, is always greater than or equal to 0.<br>
So this case tells us that, for x > 0, every value is a solution.  So one part of the solution set is (0, infinity).<br>
(2) IF x < 0, THEN  (the direction of the inequality changes)
{{{x^2 + 9 <= 6x}}}
{{{x^2-6x+9<= 0}}}
{{{(x-3)^2 <= 0}}}<br>
But again any expression, when squared, is greater than or equal to 0.  So the only POTENTIAL solution from this case is x=3 -- but this case only applies to negative values of x.  So this case adds nothing to the solution set.  In other words, it confirms the informal analysis I showed near the beginning of my response, where informal analysis showed there would be no solutions for negative values of x.<br>
And so, in the end, the complete solution set is (0, infinity).