While practicing math for the SAT, I came across the question-

What is the sum of the solutions to $\sqrt{3x+13} = x+3$?

The first step I did was to make it a quadratic from the given equation.

Working with the quadratic, $x^2 + 3x -4$, I factored it down to $(x+4)(x-1)$.

I then checked each solution with the original equation. For $x=1$, I found the equation to hold true as $\sqrt{(3*1)+13} = 1+3$

My question then comes from when I tried to check $x=-4$.

$\sqrt{(3*-4)+13} = -4+3$

$\sqrt{(-12)+13} = -1$

$\sqrt{1} = -1$ (Or so I thought)

The answer to the problem was 1 as it did not include $x=-4$ to be a proper solution. Upon seeing this, I decided to do some research.

According to the book, Algebra, by I.M. Gelfand, Alexander Shen:

"To be exact, a square root of a nonnegative number $a$ is a nonnegative number whose square is equal to $a$"

For example: $\sqrt{25} = 5$ and $5^2 = 25$

What confuses me is why we cannot also include $-5$ as a solution to the square root, because $-5^2 = 25$


Mathematicians are free to adopt any definitions. But it is customary to consider the arithmetic square root of nonnegative numbers, with the root itself being nonnegative.

This makes $f(x)=\sqrt{x}$ a (single-valued) function, defined for all real $$x\ge0.$$

(So one answer to the question why? is: because we want to make $\sqrt{x}$ a function.)

When we want the negative counterpart to $\sqrt{a}$, we write: $-\sqrt{a}$.


The negative counterpart of a square root is ALWAYS considered in a solution.

But the $\sqrt{3x + 13}$ isn't part of the solution it's part of the question.

Okay. So "the" square root is always positive. This is just a convention so that we can tell the two square roots apart. And both the positive and negative square root will matter in finding solutions.


this problem is

$\sqrt{3x + 13} = x + 3$ has the square root in the statement of the problem. It is positive because we say it is positive. If the question were $-\sqrt{3x + 13} = x+3$ (or equivalently $\sqrt{3x + 13} = -x - 3$) we'd only consider negative values. ... because that is what the problem asks for.

So we want to solve ${3x + 13} = (x+3)^2$ WITH the stipulation that $x + 3 \ge 0$.

We solve it: $x^2 + 6x +9 = 3x + 13$ so $x^2 + 3x- 4$ so solutions are among $-4$ and $1$. But $-4 + 3 < 0$ so although it is a solution to $\pm\sqrt{3x+13} =x + 3$ and it is a solution to $-\sqrt{3x+13} = x+3$, it is not a solution to $\sqrt{3x + 13} = x+3$ because... that is what the question asked for.

This is a tiny bit like asking "name all boys name that are begin with A or higher and are Germanic in origin", then getting a list of all boys that are germanic and listing them. Then finding out those that begin with A weren't listed in the solution, and then wondering why we can't include names beginning with A?


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