How do I solve this quadratic inequality with fractions?

Question

I'm a little confused with quadratic inequalities? Can someone explain how do I work out the final exam? I can do the factorising, etc but I'm not too sure how to determine ALL values of x.

eg.

$\frac{x+4} {2x-4} ≤ x - 2$

Doing the simplifying if $ x > 0$

I end up with

$0 ≤ (x-4)(2x -1)$

so, points of interest are:

$x = 1/2 $

$x = 4 $

How do I work out all values that $x$ is pointing to? I tried testing points right of the $x=4$ on the number line.

eg. $x = 5$

$ 0 ≤ (5-4)(2(5) -1)$

ie. $0 < 9$

so - it looks like its true $x > 4$

Test left of $1/2$:

$ 0 ≤ (0-4)(0-1) $

ie. $ 0 < 4$

so, its true again. What I don't understand why we perform the testing step? Is it to test if the region is actually matching the inequality?

Also, what's the best method to find ALL values of x?

Answer 1

Here's the Wavy Curve Method:

It involves the following steps:

1. Make RHS $0$.
2. Factor the polynomials.
3. Make the coefficient of the variable of all factors positive.
4. Multiply/divide both sides of the inequality by $-1$ to remove the '$-$' sign.
5. Find the roots and asymptotes of the inequality by equating each factor to $0$.
6. Plot the points on a number line and start with the largest factor. Here the curve from the positive region of the number line should intersect that point. Now look at the power of that factor and if it's odd, then we have to change the path of the curve from their respective roots while if it's even continue in the same region.

Here's how it will be in your case:

$$\frac{x+4}{2x-4}\leq x-2$$ $$\Rightarrow \frac{x+4}{2(x-2)}-(x-2)\leq0$$ $$\Rightarrow \frac{x+4-2(x-2)^2}{2(x-2)}\leq0$$ $$\Rightarrow \frac{x+4-2x^2-8+8x}{2(x-2)}\leq0$$ $$\Rightarrow \frac{-2x^2+9x-4}{2(x-2)}\leq0$$ $$\Rightarrow \frac{2x^2-9x+4}{x-2}\geq0$$ $$\Rightarrow \frac{2(x-4)(x-\frac12)}{x-2}\geq0$$ So, after factorizing, we have the roots/critical points as $4,0.5$ and $2$.
Onto step $6$:

Since we are interested in the solutions where $f(x)\geq0$ so, that'd be the region with the '$+$' sign. $$\text{Thus, }\bf x\in \left[\frac12,2\right)\cup[4,∞) $$
Note that we don't include 2 in the solution as for that value the function wouldn't be defined.

If you're interested in knowing more about the Wavy Curve Method then check this video to know its application and this video to know the logic behind this method.

Answer 2

You have$$\frac{x+4}{2x-4}\leqslant x-2\iff\frac{-2x^2+9x-4}{2(x-2)}\leqslant0.\tag1\label A$$Note that the roots of the numerator are $\frac12$ and $4$ and that therefore the numerator is greater than or equal to $0$ in $\left[\frac12,4\right]$ and it is smaller than or equal to $0$ in $\left(-\infty,\frac12\right]\cup[4,\infty)$. On the other hand, the denominator is greater than $0$ on $(2,\infty)$ and smallar than $0$ on $(-\infty,2)$. Therefore, the RHS of $\eqref A$ holds if and only if$$x\in\left[\frac12,2\right)\cup[4,\infty).$$