#### Answer

$\sqrt 7|x-1|$

#### Work Step by Step

Simplify . $f(x)=\sqrt{7x^2-14x+7}$
Which can be written as: $\sqrt{7x^2-14x+7}=\sqrt{7(x^2-2x+1)}$
or, $\sqrt{7(x^2-2(x)(1)+1^2)}$
or, $=\sqrt{7(x-1)^2}$
As per definition of square root property,we have $(p)^{\frac{m}{n}}=(\sqrt[n] {p})^m$
Thus, $=\sqrt 7 \sqrt{(x-1)^2}$
As per definition of square root property,we have $\sqrt {x^2}=|x|$
Hence, the above exponent in radical form can be written as: $\sqrt 7|x-1|$