Please help me solve the inequality$ |x^2 - 7 x + 12| < | x - 4 |$

Question

Here's my attempt. Is it correct? Is there a better way to solve it?

Answer 1

Another way: note that $|x^2-7x+12|=|x-4||x-3|$. Then $$|x-4||x-3|<|x-4|$$ $$\frac{|x-4||x-3|}{|x-4|}<1$$ Clearly, $ x = 4 $ does not satisfy the inequality, so we can avoid it (cancel the respective terms) and this gives us $$|x-3|<1$$ that is $2<x<4$.

Answer 2

Write your inequality in the form $$|x-3||x-4|<|x-4|$$

Answer 3

Taking @pointguard0 as a starting point:

$ |x^2 - 7 x + 12| < | x - 4 |$ is the same as $ |(x-3)(x-4)| < | x - 4 |$

If $x>4$, you can get rid of all the absolute values and write $ (x-3)(x-4) < (x - 4) $, thus $(x-3) < 0$ (which never happens)

If $x \in (3, 4)$, we get $ -(x-3)(x-4) < -(x - 4) $, or, simplified, $(x-3) < 1$, which always happens. The sign is inverted because (x-4) is negative

Finally, if x<3, what we get is $(x-3)(x-4) < -(x - 4) $, which simplifies to $(x-3) > -1$, in short, $x>2$

The statement is obviously false for x=4 and true for x=3. In conclusion, it is true if $x \in (2,4)