# Why are these 2 algebraic expressions equivalent?

I just solved a long problem for my physics w/calculus homework that required a simplification using a quadratic formula. The "textbook" (flipItPhysics) came up with a different simplification than mine but it turns out they are equivalent. I can't, for the life of me, figure out how to simplify mine into theirs.

## Mine:

$$x = \frac{d\left(q\pm\sqrt{Qq}\right)}{q - Q}$$

## Theirs:

$$x = d \cdot \left(\frac{q}{q- Q}\right) \left(1 \pm\sqrt{\frac{Q}{q}}\right)$$

Can someone show me how to get from mine to theirs? I'm specifically confused about how to make the inside of the square root a division instead of a multiplication.

• You can get $\pm$ with \pm. – Brian M. Scott Jan 17 '16 at 3:54
• Ahh, thanks. I looked up Latex markup just for this post and couldn't find it. – Josue Valverde Jan 17 '16 at 4:18
• A good strategy to see how an expression may have been simplified a certain way is to "undo" the simplification. Here for example multiply the expression in the parentheses on the right with the numerator on the left and bring the $q$ inside the radical. – ThisIsNotAnId Jan 26 '16 at 4:42
• Just saw the last part of your question, here's another handy trick: $q\sqrt{\frac{Q}q} = \sqrt{q^2}\sqrt{\frac{Q}q} = \sqrt{q^2\frac{Q}q} = \sqrt{Qq}.$ You have to keep in mind that here $\sqrt{q^2}$ is a placeholder for $q$ and cannot be replaced with $-q$ as would be otherwise possible. – ThisIsNotAnId Jan 26 '16 at 5:49

Notice that both expressions have a factor of $d$ (you have it in the numerator, and the textbook has it as the first factor), and both expressions have a factor of $q-Q$ in the denominator. So let's ignore those and compare what's left. We need to show that $$q \pm \sqrt{Qq}$$ and $$q \cdot \left(1 \pm \sqrt{\frac{Q}{q}} \right)$$ are equal.

It should be obvious that if you distribute the $q$ in the second expression through the parentheses, you end up with the $q$ in the first term of the first expression. Really, then, the only things we have to compare are $\sqrt{Qq}$ and $q\sqrt{\frac{Q}{q}}$. They both contain a factor of $\sqrt{Q}$, so let's ignore that. The only real mystery is, why is $\sqrt{q}$ equal to $q\sqrt{\frac{1}{q}}$?

This turns out to be one of those properties that's really simple to understand once you see it a few times, but for some reason it is rarely taught in high school. It comes down to the fact that $$q\sqrt{\frac{1}{q}}=\frac{q}{\sqrt{q}}=\frac{\sqrt{q}\sqrt{q}}{\sqrt{q}}=\sqrt{q}$$

I think the real question here is why the textbook writes the answer in what is (in my opinion at least) less simplified than the form of the answer you came up with.

• Yu can only take the $Q$ out if you know $Q$ is positive (or know complex numbers.) I think the real key is that $q=\pm \sqrt{q^2}$. – Thomas Andrews Jan 17 '16 at 4:10
• Thanks for the detailed response! – Josue Valverde Jan 17 '16 at 23:16

It’s a little easier to go the other way: assuming that $q>0$, we have

$$q\sqrt{\frac{Q}q}=\sqrt{q^2}\cdot\sqrt{\frac{Q}q}=\sqrt{q^2\cdot\frac{Q}q}=\sqrt{qQ}\;.$$

If $q<0$, $q=-\sqrt{q^2}$, so you get $-\sqrt{qQ}$; since you have a $\pm$ sign, it doesn’t matter: as long as $q\ne 0$, you get both signs.

• Ah I see, thanks for the nice simplified response! – Josue Valverde Jan 17 '16 at 23:16
• @Josue: You’re welcome! – Brian M. Scott Jan 17 '16 at 23:17