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Given a series of inequalities which all had the variable $x$ in the denominator, the challenge is to solve them by multiplying through with the square of the denominator so as to avoid having to examine both cases of $\frac{a}{b}>0$ i.e. $\frac{a}{b}$ and $\frac{-a}{-b}$

The way it worked for the first 3 questions was that the rational inequality ends up in the form of a quadratic and then it is a simple matter of factoring to produce the two solutions for $x$.

For example: $$\frac{1}{x} \geqslant 2$$ multiplied by $x^2$ on both sides: $$x \geqslant 2x^2$$ minus $2x^2$ on both sides: $$x-2x^2 \geqslant 0$$ factoring out the x from each term: $$x(1-2x) \geqslant 0$$ invoking the null factor law and enforcing the domain restriction $x \neq 0$ to calculate the solutions to x which are $x \gt 0$ and $x \leqslant \frac{1}{2}$ (in the second case the inequality symbol has been reversed because $1$ was divided by a negative number $(-2)$ when working out the result. Upon examination of the graph, this appears to be correct:

reciprocal of x minus two and it's inequality

But for the last question: $$4+\frac{3}{x} \geqslant 0$$ by multiplying both sides by the square of the denominator and simplifying, I arrive at a quadratic which when factored looks like this: $$x(4x+3)\geqslant 0$$ which is correct, however when I apply the Null Factor Law and enforce the domain restriction $x \neq 0$ I get $x \gt 0$ and $x \geqslant -\frac{3}{4}$

However the second solution of x is clearly wrong, for when I examine the graph of the reciprocal on the Left Hand Side of the original inequality, y equals or exceeds zero for $x \leqslant -\frac{3}{4}$.

When I worked out that second $x$ I did this: $$4x+3 \geqslant 0$$ Take 3 from both sides: $$4x \geqslant -3$$ Divide both sides by four: $$x \geqslant -\frac{3}{4}$$

I haven't been working with inequalities for very long and so far my understanding is that the direction of the symbol changes when both sides are multiplied by a negative number, so I'm confused about how to process this algebraically since multiplication by a negative doesn't appear to come into it...

My intention is to solve this by multiplying both sides by the square of the denominator and without looking at the graph... am I trying to square the proverbial? Via the method I am using, it is perhaps necessary to look at the graph and consider the solutions to $x$ as mere critical points/lines?

Picture included for clarity.

4+3/x

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    $\begingroup$ $x(4x+3)\ge0$ means $x\ge0$ and $4x+3\ge0$, or $x\lt0$ and $4x+3<0$ $\endgroup$ Commented Jan 30, 2023 at 0:55
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    $\begingroup$ actually I meant or $x\le0$ and $4x+3\le0$ $\endgroup$ Commented Jan 30, 2023 at 1:03
  • $\begingroup$ I think the issue here is that, while you can be certain $x^2 \ge 0$, when you factor it apart, you can't be certain that you've factored into the negative or positive factors. To me, the better method for doing this is to take the reciprocal, which always changes the direction of the inequality. So $4+3/x \ge 0 \implies 3/x \ge -4 \implies x/3 \le -1/4 \implies x \le -3/4$. But given the discontinuity at $x=0$, you still have to examine the positive case, but it's elementary that if $x$ is positive, $3/x$ is also positive. $\endgroup$ Commented Jan 30, 2023 at 1:08

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to examine both cases of $\frac{a}{b}>0$ i.e. $\frac{a}{b}$ and $\frac{-a}{-b}$

Side note: writing -a doesn't indicate that $a$ is negative, even as $a$ and $-a$ always have opposite signs for nonzero $a.$

$$4+\frac{3}{x} \geqslant 0$$ $$x(4x+3)\geqslant 0$$ when I apply the Null Factor Law and enforce the domain restriction $x {\neq} 0,$ I get $x \gt 0$ and ${x\ge-\frac{3}{4}}$

the second solution of x is clearly wrong. Via the method I am using, it is perhaps necessary to look at the graph and consider the solutions to $x$ as mere critical points/lines?

No need to refer to any graph. The inequality $$x(4x+3)\geqslant 0$$ doesn't mean $$x(4x+3)> 0 \quad \text{and}\quad x=0,$$ but rather $$x(4x+3)> 0 \quad \text{or}\quad x=0.$$ As noted, $x\ne0,$ so $$x(4x+3)> 0\\x \quad\text{and}\quad4x+3\quad\text{are either both positive OR both negative}\\\left(x>0\quad\text{and}\quad x>-\frac34\right)\quad\text{OR}\quad\left(x<0\quad\text{and}\quad x<-\frac34\right)\\x>0\quad\text{OR}\quad x<-\frac34,$$ as required.

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  • $\begingroup$ Thanks ryang, but could you show how you went from $$x(4x+3)> 0$$ to $$x<-\frac34\quad$$? It's the reversal of the sign from greater than to less than that's perplexing me. I can see it's true, i just can't see how to make the transformation algebraically. $\endgroup$
    – duckegg
    Commented Jan 31, 2023 at 20:11
  • $\begingroup$ I understand how $𝑥$ and $4𝑥+3$ are either both positive OR both negative but without creating a table of signs or looking at the graph, is it possible know which way to turn the inequality symbol? $\endgroup$
    – duckegg
    Commented Jan 31, 2023 at 21:06

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