# Finding domain of square root expression

I have a question that states: Simplify and state the domain of

(i) $\left(\sqrt{6x-7}\right)^2$

(ii) $\sqrt{\left(6x-7\right)^2}$

I have to explain why the answers are different too, and I just can't get an answer that makes sense! I'm terrible at math and I feel so stupid!

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Related: math.stackexchange.com/questions/64165/…. I think the question is ill-posed so don't blame yourself for misunderstanding it. Also, I think you mistakenly entered the same function for both parts (i) and (ii). – Dan Brumleve Sep 15 '11 at 6:27
In a real number setting, the square root function is defined only for nonnegative arguments. (1) For what values of $x$ is $6x-7$ not negative? (2) Is $(2x-7)^2$ ever negative? – anon Sep 15 '11 at 6:29
My guess, coming from his other question, is that the first is $\Bigl(\sqrt{6x - 7}\Bigr)^2$ and the second is $\sqrt{(6x - 7)^2}$. – Dylan Moreland Sep 15 '11 at 6:30
Nah, I definately copied it right... When I went to simplify the first function I ended up with the same answer that I got for the second function (If I am right, that is [that because the function is being squared within the square root, it simply becomes 6x-7?) - there is a third part to this question that says to explain why they are different and if they are the same, to check out absolute values and I don't understand how that's relevant... Thanks for the reply though! – UVic Student Sep 15 '11 at 6:35
@UVic, notice that, for example, $\sqrt{(-3)^2}=\sqrt9=3\ne-3$, and you may see where your mistake is and what's really going on here. – Gerry Myerson Sep 15 '11 at 6:40

Let us observe some simpler case: $(\sqrt{x})^2$ and $\sqrt{(x)^2}$.

For first expression we have to find root of $x$ first and then square that result. Since the square root is defined only for nonnegative real numbers first domain is $x\in [0,+\infty)$ and $(\sqrt{x})^2 = x$.

For second expression we have to square $x$ first and then find root of that result. Since square is defined for all real numbers and always gives nonnegative result second domain is $x\in(-\infty,+\infty)$ and $\sqrt{(x)^2} = |x|$.

Now we can apply same reasoning on your specific case and result is:

• for $\left(\sqrt{(6x-7)}\right)^2$, domain is $[\frac{7}{6},+\infty)$ and the expression is $6x-7$.

• for $\sqrt{(6x-7)^2}$, domain is $x\in (-\infty,+\infty)$ and expression is $|6x-7|$.

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Why are you posting everything in a \gathered environment? – Arturo Magidin Sep 17 '11 at 5:45
@ArturoMagidin,Is that forbidden? – pedja Sep 17 '11 at 5:47
(i) It looks ugly, with every line centered; (ii) it makes it hard to edit. In any case, "because it is not forbidden" does not seem like a neighborly way to behave. Note: $x^2$ is not always positive, by the way. – Arturo Magidin Sep 17 '11 at 5:50