Square root positive/negative

Question

What is the square root of $9$? Is it always $\pm 3$ or just positive $3$?

Trying to find the solution set of this equation : $x-3 = \sqrt{x+3}$

I want to understand the concept of square root to solve the problem.

Thanks

Answer 1

There are two numbers $x$ such that $x^2=9$ : $x=3$ and $x=-3$, but with the symbol $\sqrt{9}$ we indicate only the positive one, so $\sqrt{9}=3$.

In other words, the symbol $\sqrt{x}$ indicates, by definition, the positive number $y$ such that $y^2=x$.

This implies that if we search the solution of the equation $x-3=\sqrt{x+3}$, than we must have $x-3\ge0$ and, if squaring we find a solution such that $x-3<0$ this is not a solution of the given equation.

We can say that the equation $x-3=\sqrt{x+3}$ is equivalent to the system: $$ \begin{cases} x-3\ge0\\(x-3)^2=x+3 \end{cases} $$

Answer 2

By definition $\sqrt{9}=3$ but we have a different case when we are trying to solve an equation like:

$$x^2=9$$

On that case we have as a solution the numbers $3$ and $-3$ because if we square any of them we get $9$. So the right way to solve the above problem is:

$$x^2=9 → x= \pm \sqrt{9}= \pm 3$$

For your specific case, one way is square both sides and get:

$$(x-3)^2=x+3 → x^2 -7x +6=0$$

And solve the quadratic equation in the regular way and get the solutions $1$ and $6$. Just remember that solution must respect $x-3 \ge 0$ and $x+3 \ge 0$ so just the solution $6$ is good.

P.S.: we must have $x+3 \ge 0$ because otherwise $\sqrt{x+3}$ woudn't have a real number. Remember that, for example, $\sqrt{-1}$ is not a real number. And also $x-3 \ge 0$ because it is equal to $\sqrt{x+3}$ and by definition $\sqrt{x+3} \ge 0$.

Another way to find the right solution is test the values the you found in the original equation. That is equivalent to find the restrictions that I mentioned.

Answer 3

Well, let us start with something we all know and love, the quadratic formula:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

And I have a question for you. Why is it that there is a $\pm$ in front of the square root? Isn't it already the case that we have things like $\sqrt9=\pm3$? And the answer is no. By definition, we have it that $\sqrt x\ge0$, and thus, $\sqrt9=3$ since $-3<0$.

Also notice that $\sqrt x$ is undefined for $x<0$, so by looking at the problem, we can immediately deduce that

$$x-3=\sqrt{x+3}\ge0\implies x\ge3$$

$$\sqrt{x+3}\implies x+3\ge0\implies x\ge-3$$

Now, upon squaring the original equality, we have

$$x^2-6x+9=(x-3)^2=(\sqrt{x+3})^2=x+3$$

$$x^2-7x+6=0$$

$$(x-6)(x-1)=0$$

$$\implies x=1,6$$

Checking this, only the second value is $\ge3$ and checking them in the square roots, we find that

$$-2=1-3\ne\sqrt{1+3}=2$$

$$3=6-3=\sqrt{6+3}=3$$