Squaring both sides when units are different? Given $((9) \text{inches})^{1/2} = ((0.25) \text{yards})^{1/2}$, then which of the following statements is true?


*

*$((3) \text{inches}) = ((0.5) \text{yards})$

*$((9) \text{inches}) = ((1.5) \text{yards})$

*$((9) \text{inches}) = ((0.25) \text{yards})$

*$((81) \text{inches}) = ((0.0625) \text{yards})$



My question is : Can I apply here as $x^{1/2}=y^{1/2}$ then square both sides $\implies x=y$. But as given units are different. So, 

Can you explain it, please?

 A: If you have a square of side $9 \;\text{in}$, then the area of
the square is $(9 \;\text{in})^2 = 81 \;\text{in}^2$.
This is because when the sides of two squares are in the ratio
$9:1$, their areas are in the ratio $81:1$, and we have defined
the units $\text{in}^2$ so that $1\;\text{in}^2$
is the area of a square of side $1\;\text{in}$.
From this you might get the idea that units of measurement are just
another algebraic quantity that is multiplied by the number on the left
of the units. And indeed they do seem to work that way,
because we have defined them so that they do.
So if we have a quantity $ab$, with numeric value $a$ and units $b$, then
when it make sense to square this quantity (such as when taking the area
of a square of side $ab$), the result will be $(ab)^2 = a^2 b^2$.
This serves well when different units of the same dimension occur.
For example, $3\;\text{inch} = 0.25\;\text{foot}$, so a square of side
$3\;\text{inch}$ is also a square of side $0.25\;\text{foot}$, and
its area is $9\;\text{inch}^2 = 0.0625\;\text{foot}^2$.
It appears that we can reverse this process by taking a square root,
that is, $9\;\text{inch}^2 = 0.0625\;\text{foot}^2$ is the area of a
square of side $3\;\text{inch} = 0.25\;\text{foot}$, so we might think that
$(9\;\text{inch}^2)^{1/2} = 3\;\text{inch}$
and $(0.0625\;\text{foot}^2)^{1/2} = 0.25\;\text{foot}$.
If you suppose that it is possible for something to be
measured in units of $\text{inch}^{1/2}$, and you continue to treat
it as just another algebraic quantity, then it might appear to make sense
to apply the formula $(ab)^{1/2} = a^{1/2} b^{1/2}$, so that
$(9\;\text{inch})^{1/2} = 3\; \text{inch}^{1/2}$, and that you can
therefore get back the original $9\;\text{inch}$
by squaring $(9\;\text{inch})^{1/2}$.
In that case, if $(9\;\text{inch})^{1/2}$ and $(0.25\;\text{yard})^{1/2}$
are two measurements of the exact same quantity,
we should find that their squares also measure the same quantity
as each other, so $9\;\text{inch} = 0.25\;\text{yard}$.
And it happens that this last equation is true according to the
real-life definitions of inch and yard.
But there is nothing (at least, nothing that you could reasonably be expected to know about) that is customarily measured in units
of $\text{inch}^{1/2}$, so the idea that you can take the square
root of $9\;\text{inch}$, including the units, 
is nonsensical in this setting.
It is true in a formal sense, but that sense of the word
"formal" means we manipulate the forms of things without
thinking about what they are really supposed to mean.
In my opinion, the question is formal nonsense produced by someone who apparently gave little thought to its actual meaning.
We see this sometimes in my country too (and even worse examples,
where the thing you're asked to do is not merely silly, but wrong).
The only advice I can give you is to play the game, manipulate the
symbols formally even when the underlying idea is nonsense,
and hope to play well enough that you will gain the privilege to 
study from people who can give you real insight into the subject they are teaching.

Update: From the comments on the question I have learned about
fracture toughness, whose units involve a non-integer power of length
(specifically, pressure times square root of length; there are units of length involved in the pressure as well, but they're integer powers so the result is a half-power no matter how you slice it).
That's interesting news to me. In the first example of the use of this
quantity in either of the places I've seen so far, the first thing we do
is to square it, converting the half-power to a whole power, but presumably
there is some other good reason to have the half-power in the first place.
There are a quite a few other places where we take the square root of
something that has units of measurement, for example, in the formula for
how long it takes to fall distance $y$ under gravitational acceleration $g$,
$t = \sqrt{2y/g}$. There if you double the distance, the time increases by
only $\sqrt2$. As far as the units are concerned, however, the length
dimension of $y$ cancels the length dimension of $g$, leaving only a
dimension of $\text{time}^2$ under the radical; and we are quite used to
taking the square root of squared units to get back the original units.
If it were customary to publish the value of $\sqrt{2/g}$ in reference books,
rather than just $g$, we would have a constant of dimension
$\text{time}/\text{length}^{1/2}$ to be multiplied by something of
dimension $\text{length}^{1/2}$. I've never seen this,
but it's possible there is some field of study I have not experienced yet
where it is done that way.
A: Hint: Given $\sqrt{9x}=\sqrt{0.25y}\\
\Longrightarrow3\sqrt{x}=0.5\sqrt{y}\\
3\sqrt{x}=\frac{\sqrt y}{2}\\
\sqrt y=6 \sqrt x\\
y=36 x$
Spoiler:

3. $((9) \text{inches}) = ((0.25) \text{yards})$

A: If you treat $\operatorname{in}$ and $\operatorname{yd}$ as if they were variables, then you can simplify
$\left( 9 \; \operatorname{in} \right)^{1/2} =
 \left( \dfrac 14 \; \operatorname{yd} \right)^{1/2}$
to
$3 \operatorname{in}^{1/2} =
 \dfrac 12 \; \operatorname{yd}^{1/2}$
then you can square both sides and get
$9 \; \operatorname{in} = \dfrac 14 \; \operatorname{yd}$
which is true. But I have no idea what $\operatorname{in^{1/2}}$ means physically. In most cases, you can treat units as if they were variables but there needs to be a physical interpretation of the units that you end up with or just becomes formal gibberish.
A: If (one thing)$^\frac12$ = (another thing)$^\frac12$, then we can square both sides to get

one thing = another thing

Here, "one thing" is "$9$ inches", and "another thing" is "$0.25$ yards" (I don't know why you added all those parentheses!). Hence

$9$ inches = $0.25$ yards

Your worry about the units is irrelevant here, because they are inside the outer parentheses. This question is a reminder that $(9$ inches$)^\frac12$ is not the same as $9^\frac12$ inches.
A: Your original equation is actually:
9 inches = .25 yards * (36 inches/yard)
(The unit conversion is implicit. When you make it explicit, it makes the units on both sides work).
When you take the square root of both sides, you get:
3 inches^.5 = .5 yards^.5 * (6 inches^.5/yards^.5)
This, while working, is kind of a meaningless equation, since the square root of an inch has no real physical meaning.
However, if you square both sides you get:
81 inches^2 = (1/16)yards^2 * (1296 inches^2/yards^2)
Since square inches and square yards have a physical representation (area), the equation not only makes sense, but gives you the conversion rate between square inches and square yards.
A: Answer 3. is correct.
If you say for linear measure
$$ 1 ft = 12 \, in, $$
then for area measure
$$ 1 ft^2 = 144 \, in^2, $$
and
$$ 1 ft^3 = 1728 \, in^3, $$
For derived units
Speed
$$ 1 ft/ \min = 12/60 = 0.2 in/\sec, $$
Rate of discharge or volume rate
$$   3600 m^3 / hour = 1 m^3/\sec, $$
and so on.
