1. Solving equations #2

Hi i need some help please on how to solve these problems. i tried each one but either cant figure out the answer/ i got an answer but im not sure if its correct.

1) Use an absolute value inequality to describe te numbers that are between -6 and 8 insclusively.

-So for this problem i believe its just asking us to write down/ translate the sentence. This is what i got.

-6 < x < 8

only thing is im not sure if thats correct or if this is what im suspose to do...

2)Factor completely. x^9/5 -4x^4/5 -5x^-1/5

for this i took the smallest exponet fraction which was -1/5 and got.

-x^-1/5 (x^-9 +4x^-4 +5)

after i did this i couldnt see anything else i could do to factor it out even more. Then again im not the best with fraction exponets.....

help would be much appreciated! thanks!

2. Originally Posted by Nismo
1) Use an absolute value inequality to describe te numbers that are between -6 and 8 insclusively.
If $a you can express that as $\left|x-\frac{b+a}{2}\right|<\frac{b-a}{2}$.

3. Originally Posted by Nismo
Hi i need some help please on how to solve these problems. i tried each one but either cant figure out the answer/ i got an answer but im not sure if its correct.

1) Use an absolute value inequality to describe te numbers that are between -6 and 8 insclusively.

-So for this problem i believe its just asking us to write down/ translate the sentence. This is what i got.

-6 < x < 8

only thing is im not sure if thats correct or if this is what im suspose to do...

2)Factor completely. x^9/5 -4x^4/5 -5x^-1/5

for this i took the smallest exponet fraction which was -1/5 and got.

-x^-1/5 (x^-9 +4x^-4 +5)

after i did this i couldnt see anything else i could do to factor it out even more. Then again im not the best with fraction exponets.....

help would be much appreciated! thanks!
For 1 your answer is incorrect. It does not use an absolute value.

It should look like $|x-a| where a and b are real numbers.
Hint let $a$ be the midpoint of the interval

For 2: you have factored incorrectly
Hint:
$\displaystyle x^{\frac{9}{5}}-4x^{\frac{4}{5}}-5x^{-\frac{1}{5}} = \frac{x^{\frac{1}{5}}}{x^{1/5}}\left( x^{\frac{9}{5}}-4x^{\frac{4}{5}}-5x^{-\frac{1}{5}}\right) =x^{-\frac{1}{5}}(x^2-4x-5)$

4. In fact, even ignoring the "absolute value" question "-6< x< 8" does NOT "describe the numbers that are between -6 and 8 inclusively. Because it does not include -6 and 8.