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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

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

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Your mistake is to take the intersection of the domain and the range, which is a nonsense.

Clearly the polynomial under the radical can reach $0$ (for $x=1$) and goes to infinity. Hence

$$[0,\infty).$$

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