Prove that for $x\in\Bbb{R}$, $|x|\lt 3\implies |x^2-2x-15|\lt 8|x+3|$.

The problem I have is:

Prove that for real numbers $x$, $|x|\lt 3\implies |x^2-2x-15|\lt 8|x+3|$.

Since there aren't really any similar examples in my book, I've been unsure how to first approach this problem and have been trying to use to find similar questions.

Do I try and divide the problem into regions, like at absolute value inequalities, using $x^2-2x-15=(x-5)(x+3)$

Or could I just simply have two cases, those being $x\ge 3$ and $x\lt3$ like at How to write an expression in an equivalent form without absolute values?

We know that $|x|<3$. Therefore $-3<x<3$, subtracting $5$ from each side gives:

$-8<x-5<-2$

And so:

$|x-5|<8$

This implies that:

$|x^2-2x-15|=|(x-5)(x+3)|=|x-5|\cdot |x+3|<8|x+3|$

As needed.

• How do you know to subtract 5 from each side? – Churning Butter Apr 5 '15 at 21:57
• It seems reasonable because I want to eliminate the $x-5$ term. – eranreches Apr 5 '15 at 21:59
• Eliminate from where? – Churning Butter Apr 5 '15 at 22:03
• We want to show that $|x-5|\cdot |x+3|<8|x+3|$. This suggests that we need to work on $|x-5|$. – eranreches Apr 5 '15 at 22:04
• Because $-8<x-5<-2<8$, implies that $-8<x<8$. – eranreches Apr 5 '15 at 22:29

Your factorisation in fact takes you very close to the result: $$|x^2-2x-15|=|x-5||x+3|\leq(|x|+5)|x+3|<(3+5)|x+3|=8|x+3|.$$ The first inequality above is just the triangle inequality and is not strict. The second inequality above is strict because $|x|<3$ implies simultaneously $|x|+5<3+5=8$ and $|x+3|>0$.

In general, if you have to work with absolute values and inequalities based on them, split them into two cases:

Case 1: $0 \leq x \lt 3$.

In this case, $|x| = x$ so:

$|x^2-2x-15|\lt 8|x+3| \iff x^2-2x-15 \lt 8(x+3)$

This can reduced down to $f(x) = x^2 -10x-39 \lt 0$.

$f'(x) = 2x -10$ which means $f'(x) > 0 \iff x \gt 5$

$\Rightarrow f(x)$ is decreasing from $0$ to $3 \Rightarrow f(3) \lt f(x) \leq f(0)=-39\lt 0$

Since $f(x) = x^2 -10x-39 \lt 0$ is true for $0 \leq x \lt 3$, we easily show that:

$$0 \leq x \lt 3 \Rightarrow x^2-2x-15 \lt 8(x+3)$$

Case 2: $-3 \lt x <0$

In this case, $|x| = -x$ so:

$|x^2-2x-15|\lt 8|x+3| \iff x^2+2x-15 \lt 8(3-x)$

Reduce this down to $g(x) = x^2 + 10x-39 \lt 0$.

$g'(x) = 2x + 10$ which means $g'(x) < 0 \iff x \leq -5$

$\Rightarrow g(x)$ is increasing from $-3$ to $0 \Rightarrow g(-3) \lt g(x) \lt g(0) = -39 < 0$.

Since $g(x) = x^2 + 10x-39 \lt 0$ is true for $-3 \lt x \lt 0$, we can easily show that:

$$-3 \lt x \lt 0 \Rightarrow x^2+2x-15 \lt 8(3-x)$$

This completes the proof. The key to this proof was that $x \geq 0 \Rightarrow |x| = x$ and $x \lt 0 \Rightarrow |x| = -x$

$|x^2-2x-15|\lt 8|x+3|$

can be written

$|(x+3)(x-5)|\lt 8|(x+3)|$

Since

$(x+3)\gt 0$ in the range $-3 \lt x \lt 3$

$(x-5)\lt 0$ in the range $-3 \lt x \lt 3$

we have that

$(x+3)(x-5)\lt 0$ in the range $-3 \lt x \lt 3$

and therefore we can rewrite

$|(x+3)(x-5)|\lt 8|(x+3)|$

to

$-(x^2-2x-15)\lt 8(x+3)$

and reorder to

$x^2+6x+9\gt 0$

and further reorder to

$(x+3)^2\gt 0$ in the range $-3 < x < 3$

which is obviously true

Implies means inclusion of sets of solutions.

Therefore, we could solve both inequalitites and show their sets of solutions are one inside the other.

For $|x|<3$, it is $(-3,3)$.

For $|x^2-2x-15|<8|x+3|$, we get $|(x-5)(x+3)|=|(x-5)(x+3)|<8|x+3|$. This is the same as $x\neq-3$ and $|x-5|<8$. This is $[5-8,5+8]=(-3,13]$.

Now we can simply check that $(-3,3)\subset(-3,13)$