# Math Help - Inequality

1. ## Inequality

Solve for x in the given inequality: $6x - x^2 \ge\;9$

My Work:

$-x^2 + 6x - 9 \ge\; 0$

$(-x + 3)(x - 3) \ge\; 0$

$-x + 3\ge\; 0\Longrightarrow -x \ge\; -3\Longrightarrow x\le\; 3$

$x - 3 \ge\; 0\Longrightarrow x \ge\; 3$

What did I do wrong?

2. Hello,

Originally Posted by R3ap3r
Solve for x in the given inequality: $6x - x^2 \ge\;9$

My Work:

$-x^2 + 6x - 9 \ge\; 0$

${\color{red}(-x + 3)(x - 3) \ge\; 0}$

$-x + 3\ge\; 0\Longrightarrow -x \ge\; -3\Longrightarrow x\le\; 3$

$x - 3 \ge\; 0\Longrightarrow x \ge\; 3$

What did I do wrong?
Until the red thing, it's OK

But you could notice that $-x^2+6x-9=-(x^2-6x+9)=-(x-3)^2$

Is it possible that $-(x-3)^2 \ge 0$ ?

3. how exactly is that wrong though? It's how it factors isn't it?

4. Originally Posted by R3ap3r
how exactly is that wrong though? It's how it factors isn't it?
No, your working was correct, but there are two problems :

- there is no solution (but x=3), so that's why you thought you were wrong

- you forgot a situation : -x+3<0 and x-3<0 (their product is positive ). It'll also yield to an impossibility, but you mustn't forget this case.

5. So the final answer is just x=3? That makes sense.

6. Yep.

7. hmmm amazing how a simple problem can trick me even when my working is correct