Inequalities

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June 2nd 2011, 11:31 PM
Punch

Inequalities

First part of the question says: Solve the inequality $\frac{(x+1)(4-x)}{(3x+1)^2}\geqslant 0$

And after solving, the answer is $-1\leqslant x\leqslant 4, x\neq \frac{1}{3}$

Then the next part of the question says: Hence deduce the solution to the inequality $\frac{(\sqrt{x}+1)(4-\sqrt{x})}{(3\sqrt{x}+1)^2}\geqslant 0$

On comparing, i replaced x in the answer to the first part with $\sqrt{x}$

so, $-1\leqslant \sqrt{x}\leqslant 4, \sqrt{x}\neq -\frac{1}{3}$

$-1\leqslant \sqrt{x}$ and $\sqrt{x}\leqslant 4$

$x\geqslant 0$ and $x\leqslant 16$

therefore, $0\leqslant x\leqslant 16$

Yes, that's right. The answer is correct but shouldn't the answer $x\neq -\frac{1}{3}$ also be dealt with?

So shouldn't the right answer be... $0\leqslant x\leqslant 16, x\neq \frac{1}{9}$ ? why is this not the answer? Please advise, thanks
June 3rd 2011, 02:40 AM
HallsofIvy

The difficulty is that your solution to the original inequality is wrong. There is no reason to say $x\ne \frac{1}{3}$ . It is $x= -\frac{1}{3}$ that would cause the left side to not exist.
June 3rd 2011, 02:46 AM
Punch

sorry that was a typo, so why shouldn't the solution include $\sqrt{x}\neq-\frac{1}{3}$

$x\neq \frac{1}{9}$
June 3rd 2011, 05:00 AM
HallsofIvy

$\sqrt{x}$ is never negative so there no need to exclude -1/3. And, if x= 1/9, then [tex]\sqrt{x}= 1/3[/itex], not -1/3.
June 3rd 2011, 06:21 PM
Stro

Problem 1

$\frac{(x+1)(4-x)}{(3x+1)^2} \geq 0$

Let's change the equation up a bit to really look at it. All we'll do is pull a -1 out of the $(4-x)$ to give $-(x-4)$ , then rearrange it:

$\frac{-(x+1)(x-4)}{(3x+1)^2} \geq 0$

So this is what we already know by looking at it:
1. We have a negative degree two polynomial expression on top
2. This expression has two real 0s so at some point x will be > 0
2. We can easily figure out our zeros now for the equation (-1 and 4)
3. Also it looks like if $x = -\frac{1}{3}$ because we cannot divide by 0, there would be no solution.

So our answer is:

$-1 \leq x \leq 4$ and $x \neq -\frac{1}{3}$

Note another way to write this solution:
$-1 \leq x < -\frac{1}{3}$ and $-\frac{1}{3} < x \leq 4$

Problem 2

Now on your next problem:

$\frac{(\sqrt{x}+1)(4-\sqrt{x})}{(3\sqrt{x}+1)^2} \geq 0$

$\frac{-(\sqrt{x}+1)(\sqrt{x}-4)}{(3\sqrt{x}+1)^2} \geq 0$

It's going to be mostly the same except for two differences:

1. Solving for our roots won't be as easy, for example:
$(\sqrt{x} - 4) = 0$

$\sqrt{x} = 4$

$x = 16$

$(\sqrt{x}+1) = 0$

$\sqrt{x} = -1$

There is no solution here, the sqrt(x) will never be negative.

No solution, so only 16 for a root.

So that was a little different but not too bad.

2. There will never be a real value for x that would create a 0 in the bottom of your expression on the right

For real numbers in x: $(3\sqrt{x}+1)^2 \neq 0$

So we don't have to worry about a "no solution" problem on the bottom.

And we understand that sqrt{x} will always remain non-negative. To explain this better, if we put a negative number in for x, we get a complex number in return. If we put a 0 in for x, we get a 0. And if we put a positive number in for x, we get a positive in return. Never a negative result.

We can form our answer now to be $\sqrt{x} \leq 4$

Or in other words, because $\sqrt{x}$ is non-negative: $0 \leq \sqrt{x} \leq 4$

And thusly we have our solution: $0 \leq x \leq 16$

Good luck you were on the right track it just looks like some typos threw you off.
June 3rd 2011, 07:54 PM
Prove It

$\displaystyle \frac{(x + 1)(4 - x)}{(3x + 1)^2} \geq 0$

You first need to note that $\displaystyle x \neq - \frac{1}{3}$ , because that would give a zero denominator.
Since the denominator is always nonnegative, you can multiply both sides by the denominator without changing the inequality sign

$\displaystyle (x + 1)(4 - x) \geq 0$

Quadratic inequalities are nearly always easiest solved by completing the square.

$\displaystyle \begin{align*} (x + 1)(4 - x) &\geq 0 \\ -x^2 + 3x + 4 &\geq 0 \\ -(x^2 - 3x - 4) &\geq 0 \\ -\left[x^2 - 3x + \left(-\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right)^2 - 4\right] &\geq 0 \\ -\left[\left(x - \frac{3}{2}\right)^2 - \frac{9}{4} - \frac{16}{4}\right] &\geq 0 \\ -\left[\left(x - \frac{3}{2}\right)^2 - \frac{25}{4}\right] &\geq 0 \\ -\left(x - \frac{3}{2}\right)^2 + \frac{25}{4} &\geq 0 \\ -\left(x - \frac{3}{2}\right)^2 &\geq -\frac{25}{4} \\ \left(x - \frac{3}{2}\right)^2 &\leq \frac{25}{4} \\ \left|x - \frac{3}{2}\right|^2 &\leq \frac{5}{2} \\ -\frac{5}{2} \leq x - \frac{3}{2} &\leq \frac{5}{2} \\ -1 \leq x &\leq 4\end{align*}$

Now, remembering that $\displaystyle x \neq -\frac{1}{3}$ , that means our solution is

$\displaystyle x \in \left[-1, -\frac{1}{3}\right) \cup \left(-\frac{1}{3}, 4\right]$
June 4th 2011, 01:47 AM
Archie Meade

Alternatively,
the fraction is undefined for $x=-\frac{1}{3}$

For all other x, the denominator is positive.

Then for $x\ne\ -\frac{1}{3}$

the solution reduces to $(x+1)(4-x)\ge \0$

which will be true if both factors are negative or both are positive, or either is zero.

Both are negative if $x< -1$ and $x>4$ at the same time, which is not possible.

Both are positive if $-1< x< 4$

Therefore, both factors cannot be negative and so for $x\ne\ -\frac{1}{3}$

$-1\le\ x\le\ 4$