Is $\sqrt{x^2}$ = x???

Algebra Level 2

If $1<x<4$ . Find $\sqrt{x^2-2x+1} + \sqrt{x^2-8x+16}$

3 solutions

Paul Ryan Longhas
Feb 20, 2015

$\sqrt{x^2-2x+1} + \sqrt{x^-8x+16}$ $=> (x-1) + (4-x)$ $answer => 3$ $\sqrt{x^2-8x+16} = 4 - x$ , not $x-4$ , since $x<4$ $Also, \sqrt{x^2} = |x|$

Can you explain why $\sqrt{x^2-8x+16}=4-x$ and not $x-4$ even if $x<4$ ?

Vishal Yadav - 5 years, 7 months ago

Fox To-ong
Feb 22, 2015

from the given boundaries only 2 and 3 are the best choices

Samjoe Army
Jul 30, 2017

Given expression is equal to $|x-1| + |x-4|$ . So the answer is $3$