solve $|x-6|>|x^2-5x+9|$

Question

solve $|x-6|>|x^2-5x+9|,\ \ x\in \mathbb{R}$

I have done $4$ cases.

$1.)\ x-6>x^2-5x+9\ \ ,\implies x\in \emptyset \\ 2.)\ x-6<x^2-5x+9\ \ ,\implies x\in \mathbb{R} \\ 3.)\ -(x-6)>x^2-5x+9\ ,\implies 1<x<3\\ 4.)\ (x-6)>-(x^2-5x+9),\ \implies x>3\cup x<1 $

I am confused on how I proceed.

Or if their is any other short way than making $4$ cases than I would like to know.

I have studied maths up to $12$th grade. Thanks.

Answer 1

HINT:

As $x^2-5x+9=\dfrac{(2x-5)^2+11}4\ge\dfrac{11}4>0\forall $ real $x$

So, $|x^2-5x+9|=+(x^2-5x+9)\forall $ real $x$

So, the problem reduces to two cases for $|x-6|$

Also, $|x-6|>\dfrac{11}4\implies \pm(x-6)>\dfrac{11}4$

Answer 2

It usually helps to draw a graph of the functions involved.

First of all, note that the discriminant of the parabola $P: x^2 - 5x + 9 = 0$ is $5^2 - 4\cdot9 < 0$, thus $x^2 - 5x + 9$ is always positive. This means that you are looking for the points $x \in \Bbb{R}$ for which the graph of $|x-6|$ lies above $P$.

Then observe that the line $\ell: x - 6 = 0$ intersects $P$ if and only if $x^2 - 5x + 9 = x - 6$, i.e. if and only if $$ x^2 - 6x + 15 = 0 $$ which again is a quadratic equation with negative discriminant, thus $\ell$ lies below $P$, i.e. $|x-6|$ lies below $P$ for every $x \geq 6$.

On the other hand, the line $r: 6-x = 0$ intersects $P$ in two points because $$ x^2 - 4x + 3 = 0 $$ has roots $x = 1$ and $x = 3$. This means that $r$, and thus the graph of $|x - 6|$ lies above $P$ precisely for $1 < x < 3$.

Note: Here I have tacitly used the fact that $P$ will always lie above any given line for $x$ large or small enough.

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TL;DR: When confronted with absolute values do not blindly write down all the possible cases. Usually it pays to first try to figure out when the arguments of the absolute values are positive. Also it helps to visualise things from a geometric point of view, especially when low-degree polynomial functions are involved.

Answer 3

Another way is to note the inequality is equivalent to $$(x-6)^2>(x^2-5x+9)^2\iff -(x-1)(x-3)(x^2-6x+15)>0$$ The quadratic is always positive, so this is the same as $(x-1)(x-3)<0$ which means $x\in (1,3)$.