Simplify $D=\sqrt{x^2}+\sqrt{x^2-2x+1}$

Question

I have this expression, and I have to simplify it:

$D=\sqrt{x^2}+\sqrt{x^2-2x+1}$

Answer: $1)$ If $x<0, D=1-2x$, $2)$ if $0\le x<1, D=1;$ $3)$ if $x>1, D=2x-1$.

I received $|x-1| + |x|$, but I am not sure how to continue, and I have problems with understanding the answer.

$D = \sqrt{x^2}+\sqrt{(x-1)^2}=|x| + |x-1|$

Kavi Rama Murthy · Accepted Answer · 2019-07-02T09:56:22.103

2

$\sqrt{x^{2}}=|x|$ and $\sqrt {x^{2}-2x=1}=\sqrt {(x-1)^{2}}=|x-1|$. Now use the definition of absolute value function to complete the answer.

For example, $0 \leq x \leq 1$ implies $|x|=x$ and $|x-1|=1-x$ so $D=x+(1-x)=1$.

edited Jul 02 '19 at 09:56

answered Jul 02 '19 at 09:54

Kavi Rama Murthy

I edited the question. – Valeri Jul 02 '19 at 09:54
If $x<0$, how can I write $|x-1|$? – Valeri Jul 02 '19 at 10:01
When $x <0$ we have $x-1<-1<0$ so $|x-1|=-(x-1)=1-x$. – Kavi Rama Murthy Jul 02 '19 at 10:02
When $x >1$ both $x$ and $x-1$ are positive, so $D=x+(x-1)=2x-1$. – Kavi Rama Murthy Jul 02 '19 at 10:03
But why we didn't look at $x = 1$? The answer given is for $0\leq x < 1$ – Valeri Jul 02 '19 at 10:14
1

You are right. $x=1$ is missing in the answer. Case 2) can be changed to $0 \leq x \leq 1$. – Kavi Rama Murthy Jul 02 '19 at 10:16

1 Answers1