1. ## Polynomial inequalities

I have a question I worked out the problems on the bottom but it says my answers are wrong. Can you please tell me what I did wrong?

The first problem is the first attached file for it I got the answer [2,4] by doing the sign chart

The second problem is the second file attached for it I got the answer [-4,0]U[8,13]

The third problem is the third file attached and for it I got (-infinity,-6)U(6,infinity)

The fourth problem is the fourth file attached and for that I got (-infinity,-13]U[4,infinity)

All those answers are wrong, so can you tell me what I did wrong?

2. Originally Posted by Girlaaaaaaaa
I have a question I worked out the problems on the bottom but it says my answers are wrong. Can you please tell me what I did wrong?

The first problem is the first attached file for it I got the answer [2,4] by doing the sign chart

The second problem is the second file attached for it I got the answer [-4,0]U[8,13]

The third problem is the third file attached and for it I got (-infinity,-6)U(6,infinity)

The fourth problem is the fourth file attached and for that I got (-infinity,-13]U[4,infinity)

All those answers are wrong, so can you tell me what I did wrong?
Most of your answers are wrong because you include the case that the denominator equals zero. For instance:

The solution of the 1rst problem is (2, 4). With your solution you actually divide by zero

I've got a total different solution with the 3rd problem: $\frac74 < x < 6$ or as an interval: $\left(\frac74 , 6 \right)$

3. Originally Posted by earboth
Most of your answers are wrong because you include the case that the denominator equals zero. For instance:

The solution of the 1rst problem is (2, 4). With your solution you actually divide by zero

I've got a total different solution with the 3rd problem: $\frac74 < x < 6$ or as an interval: $\left(\frac74 , 6 \right)$
I just tried your answer for the third problem and it showed that it is incorrect also. Btw how did you get that answer?

4. Originally Posted by Girlaaaaaaaa
I just tried your answer for the third problem and it showed that it is incorrect also. Btw how did you get that answer?
You are right - but I am right too :

$\frac1{x-6}>4$ . You can multiply the inequaltity by (x-6) if you pay attention to the sign of the denominator:

$\frac1{x-6}>4~\iff~1>4x-24~\wedge~x-6>0~\iff~ \frac{25}4 > x ~\wedge x>6$ that means $\left(6~,~\frac{25}4\right)$

$\frac1{x-6}>4~\iff~1<4x-24~\wedge~x-6<0~\iff~ \frac{25}4 < x ~\wedge x<6$ that means there isn't any real x which satisfies this case.

5. Hello,

For the first one, you can say that the inverse function decreases. This means that for every x & y, if 1/x < 1/y, 1/(1/x)=x > 1/(1/y)=y

So it's the same as solving x-4 > x-2, which is always false. So there is no solution...

For the third one, it's the same method. If i'm not mistaking, the answer is x<25/4, excluding x=6 because it's a zero of the denominator.

6. Here a comment about the usual way this taught in North America.
Solving $\frac{1}{{x - 6}} > 4$ students are encouraged to ‘compare to zero’.
$\frac{1}{{x - 6}} - 4 > 0 \Rightarrow \quad \frac{{25 - 4x}}{{x - 6}} > 0 \Rightarrow \quad \frac{{4x - 25}}{{x - 6}} < 0$.
It is easy to check to see what values that make the statement true.