1
$\begingroup$

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

$\endgroup$

closed as off-topic by Leucippus, C. Falcon, Adam Hughes, Claude Leibovici, user91500 Jan 1 '17 at 9:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, C. Falcon, Adam Hughes, Claude Leibovici, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Square root is multi-valued and the positive branch is usually taken to be the principle branch. $\endgroup$ – Henricus V. Dec 31 '16 at 23:00
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Can you please explain why we should respect x-3 >=0 and x+3 >= 0 ? $\endgroup$ – InquisitiveGirl Dec 31 '16 at 23:37
  • $\begingroup$ Please, see the update. $\endgroup$ – Arnaldo Jan 1 '17 at 0:38
2
$\begingroup$

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: $$ \begin{cases} x-3\ge0\\(x-3)^2=x+3 \end{cases} $$

$\endgroup$
0
$\begingroup$

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

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

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.