# If $(\pm x)^2 = X$ then $\sqrt{X} = x$?

I've been helping my siblings with their GCSE and A Level maths and I've come across a question where they have just taken the positive square root. It's a pure maths question and there's no (obvious) reason to ignore the negative square root.

I always thought that the square root always gave two values, a positive and negative, and this is shown in the quadratic formula where we have $\pm \sqrt{b^2 - 4ac}$.

So what is the correct answer?

Should you always take just the positive or the negative aswell?

In the same way, if we have $\cos(\theta) = x$, why do we not say then $\theta = \pm \cos^{-1}(x)$ as $\cos(-\theta) = \cos(\theta)$ isn't it?

EDIT: One question wants me to work out the normal of the curve $y^2 = 4ax$ where $a$ is a positive constant at the point $(at^2, 2at)$. They start by taking the square root and write $y = \sqrt{4ax}$.

In another question, they want me to work out the tangent to the line $y^2 = 27x$ at the point $(3,9)$. Here, they also start by taking the square root but write $y = \pm \sqrt{27x}$. Why have they taken the positive in one and both in the other?

It'd probably be easier to do it using implicit differentiation, but I just want to understand why the square root bit is different.

• The cause for confusion is that it's never made clear (when it should be) that "$x$ such that $x^2=y$" and "$\sqrt{y}$" don't necessarily mean the same thing. – Clive Newstead Mar 27 '13 at 16:26
• If you have good reasons to get only one value, go on. But in the general case, you must not forget solutions. For example, if $\cos \theta = x$, then $\theta = \pm \arccos x + 2k\pi$, but if you know it's an angle in a triangle, then necessarily $\theta \in [0,\pi]$ and $\theta = \arccos x$. Remember that arc cosine, as well as square root, are perfectly definite functions, with only one value. Square root maps $[0, \infty[$ to $[0, \infty[$, and arc cosine maps $[-1,1]$ to $[0,\pi]$. – Jean-Claude Arbaut Mar 27 '13 at 16:48
• Concerning your edit, in the first case if you assume $t \geq 0$, you are on the "positive" branch, so taking positive square root is sound. Likewise, in the second case, it's also assumed you are on the "positive" branch, since $x=3$ and $y=9$, so you should also take the positive square root. Maybe a typo in the question? – Jean-Claude Arbaut Mar 27 '13 at 17:13

There are always two numbers that square to a given number. By convention the symbol $\sqrt{x}$ represents the positive of the two numbers that square to $x$. The other number is given by putting in the negative sign yourself: $-\sqrt{x}$.
This is why, when manipulating equations you always put in a $\pm$ after taking a square root, because you are not sure whether the solution you seek is the positive or negative root. This is also why the formula, in general, is $\sqrt{x^2} = |x|$, instead of $=x$.
As for $\cos$, the function $\cos^{-1}$ always gives values between $0$ and $\pi$. This is again just a convention, so that the symbol $\cos^{-1}(x)$ stands for something concrete. If you wish to shift values by $2\pi$ or add in a negative sign to obtain another solution to $x = \cos(\theta)$ then you should do that yourself, but $\cos^{-1}(x)$ only stands for one of the infinitely many possible solutions to that equation, just like $\sqrt{a}$ only stands for one of the two possible solutions to $x^2 = a$.
• So really we are not thinking the square root of $X$, we are thinking what could be multiplied (or squared) to give $X$? – Kaish Mar 27 '13 at 16:32
• If you mean what could be multiplied by itself to give $x$, then that is the exact same thing as a square root of $x$. The sticking point is that I said "a" square root, because in fact there are two, so in order for $\sqrt{x}$ to have a definite meaning you have to pick one. The convention is to pick the positive one. – Jim Mar 27 '13 at 16:35