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Question : Find solution for $x \in \Bbb R$, $|x^2+6x+8|=|x^2+4x+5|+|2x+3|$

I considered 8 different cases and arrived at the answer $\big [ \frac{-3}{2}, \infty)$

I dont know if its correct. Also, considering 8 different cases is tedious. Is there any other method to solve problems like these?

Jaideep Khare
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Yami Kanashi
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  • I think you should get away with at most $6$ cases, instead of $8$, since $x^2+6x+8$ changes sign in at most two places on the number line, $x^2+4x+5$ in at most two places and $2x+3$ in one place. That's at most five sign changes, dividing the number line into at most $6$ regions. That being said, the simplification by Jaideep below is more significant. – Arthur Nov 30 '17 at 08:01
  • @arthur Exactly! i repeated few cases unknowingly. My bad. – Yami Kanashi Nov 30 '17 at 08:18
  • Note that $x^2+4x+5=(x+2)^2+1$ is always positive, so in fact you would only have $4$ cases. – Arnaud D. Nov 30 '17 at 08:49

2 Answers2

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If you observe the equality, it's of the form $$|u+v|=|u|+|v|$$

But we also have following triangle inequality $$|u+v| \le |u|+|v| \quad \forall \; u,v \in \Bbb R$$

This implies that either both $u$ and $v$ are positive, or both are negative.

Hence, only required condition is $$u \cdot v \ge 0$$

Now, since $u=x^2+4x+5=(x+2)^2+1$, it's always positive.

Therefore we need to solve for $v \ge 0$ i.e. $$(2x+3) \ge 0 \implies x \ge - \frac 32$$

Jaideep Khare
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The equation is piecewise quadratic and the pieces are delimited by the zeros of the arguments of the absolute values.

We have

$$|x^2+6x+8|=\begin{cases}x\le-4\lor x\ge-2&\to x^2+6x+8\\-4\le x\le-2&\to-(x^2+6x+8),\end{cases}$$

$$|x^2+4x+5|=x^2+4x+5,$$

$$|2x+3|=\begin{cases}x\le-\dfrac32&\to-(2x+3)\\x\ge-\dfrac32&\to2x+3.\end{cases}$$

There are actually four pieces to consider:

$$\begin{cases} x\le-4&\to x^2+6x+8=x^2+4x+5-(2x+3)\to\text{incompatible} \left(x=-\dfrac32\right)\\ -4\le x\le-2&\to-(x^2+6x+8)=x^2+4x+5-(2x+3)\to\text{no roots}\\ -2\le x\le-\dfrac32&\to x^2+6x+8=x^2+4x+5-(2x+3)\to x=-\dfrac32\\ -\dfrac32\le x&\to x^2+6x+8=x^2+4x+5+2x+3\to\text{any }x. \end{cases}$$

Note that this is a general approach. As noticed by Jaideep Khare, there is a shortcut for this particular problem.

In the case of $q$ quadratic terms with real roots and $l$ linear terms, the combinations lead to $2q+l+1$ intervals (if you handled all bounds independently, you would have to consider $3^q2^l$ cases !)