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?
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.
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.