To solve the expression [tex]\(\sqrt{a^2} - |a-c| + \sqrt{(b-c)^2}\)[/tex], we need to break it down step by step.
1. Simplify [tex]\(\sqrt{a^2}\)[/tex]:
We know that [tex]\(\sqrt{a^2}\)[/tex] equals the absolute value of [tex]\(a\)[/tex], so [tex]\(\sqrt{a^2} = |a|\)[/tex].
2. Simplify [tex]\(\sqrt{(b-c)^2}\)[/tex]:
Similarly, [tex]\(\sqrt{(b-c)^2}\)[/tex] equals the absolute value of [tex]\(b-c\)[/tex], so [tex]\(\sqrt{(b-c)^2} = |b-c|\)[/tex].
3. Combine the terms:
The expression becomes:
[tex]\[
|a| - |a-c| + |b-c|
\][/tex]
Given that the positions of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] on the number line are such that [tex]\(a\)[/tex] is to the left of [tex]\(c\)[/tex] and [tex]\(b\)[/tex] is also to the left of [tex]\(c\)[/tex], we need to consider the possible values of the absolute values.
4. Consider the absolute values:
- If [tex]\(a < c\)[/tex], then [tex]\(|a-c| = c-a\)[/tex] and [tex]\(|a| = -a\)[/tex].
- Regardless of the positions of [tex]\(b\)[/tex] in relation to [tex]\(c\)[/tex], [tex]\(|b-c| = c-b\)[/tex] if [tex]\(b < c\)[/tex] and [tex]\(|b-c| = b-c\)[/tex] if [tex]\(b > c\)[/tex]. We will assume a typical case where [tex]\(b < c\)[/tex] similar to [tex]\(a < c\)[/tex].
5. Substitute these into the equation:
Since [tex]\(a < c\)[/tex] and considering typical cases such as [tex]\(a < b < c\)[/tex], we assume:
[tex]\[
\sqrt{a^2} = |a| = -a \quad \text{(since a is negative and absolute value of a negative number)} \\
|a-c| = c - a \\
\sqrt{(b-c)^2} = |b-c| = c - b \\
\][/tex]
6. Recalculate the expression:
[tex]\[
|a| - |a-c| + |b-c| = (-a) - (c - a) + (c - b)
\][/tex]
Simplify the equation:
[tex]\[
-a - c + a + c - b
\][/tex]
7. Combine like terms:
[tex]\[
-a + a - c + c - b = -b
\][/tex]
Thus, the simplified expression results in [tex]\(-b\)[/tex]. So the answer is:
[tex]\[
\boxed{-b}
\][/tex]
