When do I have to take the solution of a square root of a number with negative sign?

Question

I saw instances where we took the answer of a square root of a number/expression with Negative and Positive sign. And also some instances where we took the answer only with positive sign? Why and in what cases does this happens? Please explain me the rule.

Answer 1

If $x$ is a real positive number, the symbol $\sqrt x$ represents the only real positive number $y$ such that $y^2=x$.

Answer 2

In general, when solving an equation of the form $$ x^2 = \text{something}, $$ the solutions are always $+\text{something}$ and $-\text{something}$ because the statement $x^2 = \text{something}$ is true exactly when $x = +\text{something}$ or when $x = -\text{something}$.

The function $\sqrt{\,\cdot\,}\,\colon \Bbb R_{\ge 0}\to\Bbb R_{\ge 0}$ is a function that takes a nonnegative real number $x$ and returns the unique nonnegative real number $y$ that satisfies $y^2 = x$.

Answer 3

One should be careful with the distinction between equations such as

$$ x = \sqrt{25} $$

and

$$ x^2 = 25 $$

The former ($x = \sqrt{25}$) states that $x$ is equal to the square root of $25$. By convention, the square root of $25$ is that non-negative value whose square is equal to $25$—namely, the number $5$. With this interpretation, $x \not= -5$.

The latter ($x^2 = 25$) is an equation that can be solved for $x$. The typical way to solve it is to "take the square root of both sides"—an operation that is fraught with peril. In truth, what we do is we find those values which, when squared, yield the original equation. If we do this, we get something like

$$ \pm x = \pm \sqrt{25} = \pm 5 $$

This is, in some sense, shorthand for "either $+x = +5$ or $+x = -5$ or $-x = +5$ or $-x = -5$". We note, then, that $+x = +5$ and $-x = -5$ are equivalent, and likewise $+x = -5$ and $-x = +5$ are equivalent, so that we can write the foregoing as

$$ x = \pm 5 $$

On the other hand, consider an equation such as

$$ \sqrt{2x^2-2x+1} = -x $$

We might straightforwardly try to solve this by squaring both sides to get

$$ 2x^2-2x+1 = x^2 $$

and then

$$ x^2-2x+1 = 0 $$

which we can rewrite as

$$ (x-1)^2 = 0 $$

We can then take the square root of both sides (!) to get

$$ x-1 = 0 $$

which of course yields $x = 1$. However, if we plug $x = 1$ into our original equation, we get

$$ 1 = \sqrt{1} = \sqrt{2(1)^2-2(1)+1} \stackrel{?}{=} -1 $$

which does not work out. In fact, the original equation has no solutions in the reals. What happened, of course, is that in squaring that original equation, we got an equation that is satisfied whenever the original equation is, but not vice versa: That is, there are situations where the squared equation is valid but the original equation isn't—namely, $x = 1$. So one has to be careful to consider the original context when taking square roots.

Answer 4

Generally when somebody writes $\sqrt x$, it's taken by convention to mean the positive square root. However, if you take the square root of both sides yourself while solving the equation, you should always use $\pm$. Sometimes you know from other parts of your problem that the negative answer is invalid; for example, when solving a projectile problem in physics, an answer which gives you a negative value for time is usually thrown out/not relevant. But it never hurts to take $\pm$ and plug in all your answers at the end to see if they work, and do a reality check to make sure they make sense within your problem.