sign of a square root

tucooooo

New member
Hi everyone,
I want to demonstrate that the equation has no fiber when x is a positive integer.
I am trying to calculate and I expect that i should get a negative square root (and thus demonstrate that it´s impossible).
But I don´t really know what to do with the negative sign that is before the square root.
Any idea?

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Subhotosh Khan

Super Moderator
Staff member
Hi everyone,
I want to demonstrate that the equation View attachment 16013 has no fiber when x is a positive integer.
I am trying to calculate View attachment 16012 and I expect that i should get a negative square root (and thus demonstrate that it´s impossible).
But I don´t really know what to do with the negative sign that is before the square root.
Any idea?
You wrote:

$$\displaystyle -\sqrt{x^2-3}$$

$$\displaystyle -\sqrt{x^2} \ \ - 3$$

Those two are NOT equivalent statements.

Dr.Peterson

Elite Member
I want to demonstrate that the equation View attachment 16013 has no fiber when x is a positive integer.
I am trying to calculate View attachment 16012 and I expect that i should get a negative square root (and thus demonstrate that it´s impossible).
But I don´t really know what to do with the negative sign that is before the square root.
What does "fiber" mean here? (And what you call an equation in the first line is an expression, not an equation.)

If you are asking how to show that the equation $$\displaystyle -\sqrt{x^2 - 3} = 2$$ has no solution, just use the fact that the square root can never be negative, so its negative can never be positive.

That is, $$\displaystyle \sqrt{x^2 - 3} \ge 0$$, so $$\displaystyle -\sqrt{x^2 - 3} \le 0$$.

Mr. Bland

Junior Member
I am trying to calculate View attachment 16012 and I expect that i should get a negative square root (and thus demonstrate that it´s impossible).
It's important to remember that square roots can be negative: squaring a negative number gives a positive result, so that negative number is a valid square root. Consider the equation $$\displaystyle x^2 - 4 = 0$$. If you graph the parabola, it visibly crosses the X axis at +2 and -2, which are the solutions for the equation and the square roots of 4.

In cases where both square roots are relevant, you'll usually see the plus/minus sign $$\displaystyle \pm$$, but it's not strictly required. In the given equation, the graph of $$\displaystyle -\pm\sqrt{x^2} - 3$$ consists of two perpendicular lines that intersect at point (0, -3). There are two places where the graph crosses the X axis: -3 and +3, but only in the case of the negative square root.

On the other hand, square roots and negative numbers do have an "impossible" relationship that crops up from time to time: taking the square root of a negative number. Wherever invalid expressions and square roots come up together, it is usually in this context. (Square roots of negative numbers is a thing in complex arithmetic, but that's another topic.)

Cubist

Junior Member
square roots can be negative
I'm not sure about that statement. I think the square root function $$\displaystyle \sqrt{x}$$ by definition returns the +ve solution only.

However, then you can say that if $$\displaystyle a=b^2$$ then $$\displaystyle b=\pm{\sqrt{a}}$$

HallsofIvy

Elite Member
It is easy to confuse two different concepts. Yes, the number "a" has two "square roots", two numbers whose square is a, $$\displaystyle \sqrt{a}$$ and $$\displaystyle -\sqrt{a}$$. The reason we need the "-" in the second is that $$\displaystyle \sqrt{a}$$ means only the positive root.

Cubist

Junior Member
I'm not sure about that statement. I think the square root function $$\displaystyle \sqrt{x}$$ by definition returns the +ve solution only.

However, then you can say that if $$\displaystyle a=b^2$$ then $$\displaystyle b=\pm{\sqrt{a}}$$
Actually I now think your statement is correct since "square root" is different to the "square root function". There are two valid square roots, and we can refer to them unambiguously as $$\displaystyle \sqrt{a}$$ and $$\displaystyle -\sqrt{a}$$ because the "square root function" always gives the positive one (like HallsofIvy explained)

I do find it a bit confusing!

Mr. Bland

Junior Member
, meaning my statement about $$\displaystyle \pm$$ being optional is incorrect. Therefore, regarding the original post, it is unambiguously true that real $$\displaystyle -\sqrt{x^2} \le 0$$ as long as real $$\displaystyle x \ge 0$$.

The takeaway, at least for me, is that if you wind up producing a radical algebraically, remember to use the $$\displaystyle \pm$$ notation where applicable:
$$\displaystyle x^2 - 5 = 0 \to x = \pm\sqrt{5}$$​

HallsofIvy

Elite Member
Research into the $$\displaystyle \sqrt{}$$ notation has turned up that it does unambiguously refer to the , meaning my statement about $$\displaystyle \pm$$ being optional is incorrect. Therefore, regarding the original post, it is unambiguously true that real $$\displaystyle -\sqrt{x^2} \le 0$$ as long as real $$\displaystyle x \ge 0$$.
It is "unambiguously true" that $$\displaystyle -\sqrt{x^2}\le 0$$ for x any real number, even negative numbers!

The takeaway, at least for me, is that if you wind up producing a radical algebraically, remember to use the $$\displaystyle \pm$$ notation where applicable:
$$\displaystyle x^2 - 5 = 0 \to x = \pm\sqrt{5}$$​

tucooooo

New member
You wrote:

$$\displaystyle -\sqrt{x^2-3}$$

$$\displaystyle -\sqrt{x^2} \ \ - 3$$

Those two are NOT equivalent statements.
Sorry, i meant the first one.

tucooooo

New member
What does "fiber" mean here? (And what you call an equation in the first line is an expression, not an equation.)

If you are asking how to show that the equation $$\displaystyle -\sqrt{x^2 - 3} = 2$$ has no solution, just use the fact that the square root can never be negative, so its negative can never be positive.

That is, $$\displaystyle \sqrt{x^2 - 3} \ge 0$$, so $$\displaystyle -\sqrt{x^2 - 3} \le 0$$.
Thanks!! I think that this is the solution. I just didn't know that rule.

Jomo

Elite Member
Hi everyone,
I want to demonstrate that the equation View attachment 16013 has no fiber when x is a positive integer.
I am trying to calculate View attachment 16012 and I expect that i should get a negative square root (and thus demonstrate that it´s impossible).But I don´t really know what to do with the negative sign that is before the square root.
Any idea?