Sum of squares of roots from absolute value equations

Expert intro: This problem combines absolute value equations with root accounting. The key is to split each equation into cases, find all real roots, and then compute the required sum of their squares.

Detailed walkthrough

Solve the first equation

Consider

$|x-2|^2+|x-2|-2=0.$

Let

$u=|x-2|\ge 0.$

Then the equation becomes

$u^2+u-2=0.$

Factor:

$(u+2)(u-1)=0.$

Since $u\ge 0$, we keep only

$u=1.$

So

$|x-2|=1,$

which gives

$x=1 \quad \text{or} \quad x=3.$

Their squares sum to

$1^2+3^2=1+9=10.$

Solve the second equation

Now consider

$x^2-2|x-3|-5=0.$

Let

$v=|x-3|.$

Then

$x^2-2v-5=0.$

We split into cases based on the sign of $x-3$.

Case 1: $x\ge 3$

Then $|x-3|=x-3$, so

$x^2-2(x-3)-5=0,$

which simplifies to

$x^2-2x+1=0,$

or

$(x-1)^2=0.$

This gives $x=1$, but it does not satisfy $x\ge 3$, so it is invalid.

Case 2: $x<3$

Then $|x-3|=3-x$, so

$x^2-2(3-x)-5=0,$

hence

$x^2+2x-11=0.$

Solve:

$x=\frac{-2\pm\sqrt{4+44}}{2}=-1\pm 2\sqrt{3}.$

Both values satisfy $x<3$, so both are valid roots.

Compute the required sum

The roots of the second equation are

$-1+2\sqrt3 \quad \text{and} \quad -1-2\sqrt3.$

Their squares are

$( -1+2\sqrt3 )^2=13-4\sqrt3,$

$( -1-2\sqrt3 )^2=13+4\sqrt3.$

So the sum of their squares is

$26.$

Add the first equation's contribution:

$10+26=36.$

Therefore, the required sum is

$\boxed{36}.$

Key idea to remember

Whenever an equation contains absolute values, the safest path is to isolate the absolute value expression, split into cases, and check any candidate roots against the original condition.

💡 Pitfall guide

A common trap is to assume that the roots from the second equation automatically satisfy the case used to derive them. For absolute value equations, every candidate must be checked against the inequality that defined the case. Another mistake is to stop after finding $u=1$ in the first equation and forget that $u=|x-2|$ leads to two x-values, not one. The sign of the absolute value changes the number of solutions.

🔄 Real-world variant

If the first equation were $|x-2|^2+|x-2|-6=0$, then setting $u=|x-2|$ would give $u^2+u-6=0$, so $u=2$ and $u=-3$. Only $u=2$ is allowed, leading to $x=0$ and $x=4$. If the second equation were $x^2-2|x-3|-1=0$, the same case-splitting method would apply, but the quadratic obtained in the $x<3$ branch would be different and could produce a different number of valid roots.

🔍 Related terms

absolute value equation, case analysis, sum of squares