Solve x^2/(x+3) < 9/(x+3): -2 < x < 1/2, x > 3

[eleventhxhour]

Find the solution set for x^2/(x+3) < 9/(x+3)

So I moved the term 9/(x+3) over to the left side and cross-multiplied the two fractions. Then, I simplified to get x^2-9 (because the x+3 cancel out across the fraction bar). I got x^2-9, which factors to (x+3)(x-3). Then, I created an interval table using these key values, ending with the answer -3 < x < 3.

However, the answer in the book states that it is -2 < x < 1/2 and x > 3. What did I do wrong? How did they get the -2 and 1/2 values?

Thanks!

 

[MarkFL]

This is how I would solve the problem:

\(\displaystyle \frac{x^2}{x+3}<\frac{9}{x+3}\)

\(\displaystyle \frac{x^2}{x+3}-\frac{9}{x+3}<0\)

\(\displaystyle \frac{(x+3)(x-3)}{x+3}<0\)

We don't want to remove the singularity here, as this will remove a critical number. We have two such critical numbers:

\(\displaystyle x\in\{-3,3\}\)

Both if which need to be excluded from the solution because of division by zero in the case of $x=-3$ and because the inequality is strict.

So, we have 3 intervals to consider:

\(\displaystyle (-\infty,-3),\,(-3,3),\,(3,\infty)\)

Pick a test value from each interval, and test the sign of the expression on the left, and keep those for which the expression is negative...what do you find?

 

[eleventhxhour]

This is how I would solve the problem:

\(\displaystyle \frac{x^2}{x+3}<\frac{9}{x+3}\)

\(\displaystyle \frac{x^2}{x+3}-\frac{9}{x+3}<0\)

\(\displaystyle \frac{(x+3)(x-3)}{x+3}<0\)

We don't want to remove the singularity here, as this will remove a critical number. We have two such critical numbers:

\(\displaystyle x\in\{-3,3\}\)

Both if which need to be excluded from the solution because of division by zero in the case of $x=-3$ and because the inequality is strict.

So, we have 3 intervals to consider:

\(\displaystyle (-\infty,-3),\,(-3,3),\,(3,\infty)\)

Pick a test value from each interval, and test the sign of the expression on the left, and keep those for which the expression is negative...what do you find?

So I tried that and got that it's negative for the intervals x<-3 and -3 < x < 3. But that's still different from the answer in the book? I'm not sure where they'd even get the -2 and 1/2 from.

 

[topsquark]

So I tried that and got that it's negative for the intervals x<-3 and -3 < x < 3. But that's still different from the answer in the book? I'm not sure where they'd even get the -2 and 1/2 from.

(shrugs) The book is wrong.

-Dan

 

[eleventhxhour]

(shrugs) The book is wrong.

-Dan

Alright, thanks!

So the book also asked to graph it. I know there'd be vertical asymptote at x = -3, but how would you figure out where the horizontal asymptote is (or if there even is one)?

 

[MarkFL]

...So the book also asked to graph it. I know there'd be vertical asymptote at x = -3, but how would you figure out where the horizontal asymptote is (or if there even is one)?

You won't have a vertical asymptote at $x=-3$, you will have a hole in the graph instead, because the singularity is removable.

It will be identical to the graph of $y=x-3$, except with the hole at $(-3,-6)$.