# Find all possible values which following expression can take $\sqrt{x^2-7x+6}$

Given expression is:

$\sqrt{x^2-7x+6}$.

Now i first find values of x for which this is a valid question by putting guy inside square root equals to greater than 0. I get $x \in[-\infty,1]\cup[6,\infty]$. Now i completed the square and i got $$\sqrt{(x-\frac{7}{2})^2-\frac{25}{4}}$$. Now from here i calculated range as x $\in$ $[0,\infty]$. Now i have taken intersection of values of obtained and i got answer to be $[0,1]\cup[6,\infty]$. but my textbook says answer to be $[0,\infty]$. Where is my mistake?

Thanks

• Comments are not for extended discussion; this conversation has been moved to chat. – Daniel Fischer Oct 28 '16 at 17:30

You want to see what are the real numbers $c$ such that the equation $$\sqrt{x^2-7x+6}=c$$ has a solution. As you observe, $c\ge0$ by definition of square root. Now we can square, getting $x^2-7x+6=c^2$ or $$x^2-7x+6-c^2=0$$ which has a solution if and only if its discriminant is $\ge0$. This amounts to $$49-4(6-c^2)\ge0$$ which is true for every $c$. So the range is $[0,\infty)$.
With completion of the square it's the same: the equation is $$\left(x-\frac{7}{2}\right)^{\!2}=c^2+\frac{25}{4}$$ which is solvable for every $c$.
You are right that the function is only defined over $(-\infty,1]\cup[6,\infty)$, but this doesn't affect the range.
Clearly the polynomial under the radical can reach $0$ (for $x=1$) and goes to infinity. Hence
$$[0,\infty).$$