Solving quadratic inequalities

Question

I have a quadratic inequality I am halfway through solving but cannot quite figure out the concept. $$3+\frac{4-x}x>0$$ I so far understand that without knowing whether x is positive, I would need to square it, but I do not know what comes next.

Answer 1

We have

$$3+\frac{4-x}{x} >0$$

$$\frac{3x+4-x}{x} >0$$

$$\frac{2x+4}{x} >0$$

Divide both sides by $2$

$$\frac{x+2}{x} >0$$

Now use wavy curve method

$-\infty+++++ (-2)-----(0)++++++ \infty$

As we want L.H.S. to be positive, hence we get

$x \in (-\infty, -2) \cup (0, \infty)$

Edit 1

Let me explain you how I decided the signs.

When $x>0$, then $x+2>0$ and $x>0$, hence $\frac{x+2}{x}$ will be positive.

Again when $x<-2$, then $x+2<0$ and $x<0$, hence $\frac{x+2}{x}$ will be positive.

When $-2<x<0$, then $x+2>0$ but $x<0$, hence $\frac{x+2}{x}$ will be negative.

That is how I decided the sign.

Answer 2

Hint:

I suppose that your inequality is: $$ 3+\frac{4-x}{x}>0 $$ that, for $x\ne 0$ becomes

$$ \frac{2x+4}{x}>0 $$

now note that a fraction is positive if the numerator and the denominator have the same sign.

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The inequality is equivalent to the two systems:

$$ \begin{cases} 2x+4>0\\ x>0 \end{cases} \quad \mbox{or} \quad \begin{cases} 2x+4<0\\ x<0 \end{cases} $$ with solution:

$x<-2$ or $x>0$

Answer 3

We want to find all Solutions to $3 + \frac{4-x}{x} >0$.

At first, we have to make a distinction between (i) $x>0$ and (ii) $x<0$. Since $x=0$ is not possible, all possible values of $x$ are covered by this distinction.

(i) We multiply the inequality by $x$. Since $x$ is positive, the ">" remains a ">":

$$3x + 4 - x > 0 \Longleftrightarrow 2x > -4 \Longleftrightarrow x>-2.$$

(ii) We multiply the inequality by $x$. Since $x$ is negative, this time the ">" changes to a "<":

$$3x + 4 - x < 0 \Longleftrightarrow 2x < -4 \Longleftrightarrow x<-2.$$

Conclusion: The set "L" of all solutions can be written as the Union

$$L = \{x>0, x>-2\} \cup \{x<0,x<-2\} = \{x>0\} \cup \{x<-2\} = (-\infty,-2) \cup (0, \infty).$$

Answer 4

This is not a quadratic inequality but rather a reciprocal one, as we can shuffle the left-hand side a little: $$3+\frac{4-x}x>0$$ $$3+\frac4x-1>0^*$$ $$2+\frac4x>0$$ $$\frac4x>-2$$ $$\frac1x>-\frac12$$ $$x<-2\text{ or }x>0$$ *Because we take $\frac xx=1$, we must exclude the $x=0$ case because it results in an indeterminate form.

Answer 5

We can use the following

$\frac{A}{B}>0$ and $B\not=0$ if and only if $AB>0$.

It holds for the other remaining inequality signs ($<$, $\leq$, etc). Trivially, it also holds for $=$.

So after obtaining the following,

$$\frac{x+2}{x}>0$$

where $x\not=0$, we can rewrite it as

$$x(x+2)>0$$

As the other answer called it as "wavy curve" method, draw a number line with the zeros of $x(x+2)$ and determine the sign of $x(x+2)$ at each interval.

Finally select the intervals with positive signs (because of $>$ in $\frac{x+2}{x}>0$).

Thus the solution is

$$x<-2\text{ or }x>0$$