sign of a square root

[tucooooo]

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?
thanks in advance..

 

[Deleted member 4993]

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?
thanks in advance..

You wrote:

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

But your attachment shows:

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

Those two are NOT equivalent statements.

 

[Dr.Peterson]

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 [MATH]-\sqrt{x^2 - 3} = 2[/MATH] has no solution, just use the fact that the square root can never be negative, so its negative can never be positive.

That is, [MATH]\sqrt{x^2 - 3} \ge 0[/MATH], so [MATH]-\sqrt{x^2 - 3} \le 0[/MATH].

 

[Mr. Bland]

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 [MATH]x^2 - 4 = 0[/MATH]. 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 [MATH]\pm[/MATH], but it's not strictly required. In the given equation, the graph of [MATH]-\pm\sqrt{x^2} - 3[/MATH] 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]

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]

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]

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]

Research into the [MATH]\sqrt{}[/MATH] notation has turned up that it does unambiguously refer to the principal square root, meaning my statement about [MATH]\pm[/MATH] being optional is incorrect. Therefore, regarding the original post, it is unambiguously true that real [MATH]-\sqrt{x^2} \le 0[/MATH] as long as real [MATH]x \ge 0[/MATH].

The takeaway, at least for me, is that if you wind up producing a radical algebraically, remember to use the [MATH]\pm[/MATH] notation where applicable:

[MATH]x^2 - 5 = 0 \to x = \pm\sqrt{5}[/MATH]​

 

[HallsofIvy]

Research into the [MATH]\sqrt{}[/MATH] notation has turned up that it does unambiguously refer to the principal square root, meaning my statement about [MATH]\pm[/MATH] being optional is incorrect. Therefore, regarding the original post, it is unambiguously true that real [MATH]-\sqrt{x^2} \le 0[/MATH] as long as real [MATH]x \ge 0[/MATH].

It is "unambiguously true" that $$-\sqrt{x^2}\le 0$$ 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 [MATH]\pm[/MATH] notation where applicable:

[MATH]x^2 - 5 = 0 \to x = \pm\sqrt{5}[/MATH]​

 

[Mr. Bland]

True, heh, I was thinking [MATH]\sqrt{x}[/MATH]. At least what I said is nonetheless accurate!

 

[tucooooo]

You wrote:

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

But your attachment shows:

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

Those two are NOT equivalent statements.

Sorry, i meant the first one.

 

[tucooooo]

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 [MATH]-\sqrt{x^2 - 3} = 2[/MATH] has no solution, just use the fact that the square root can never be negative, so its negative can never be positive.

That is, [MATH]\sqrt{x^2 - 3} \ge 0[/MATH], so [MATH]-\sqrt{x^2 - 3} \le 0[/MATH].

Thanks!! I think that this is the solution. I just didn't know that rule.

 

[Dr.Peterson]

So, what does it mean to say "the equation [MATH]-\sqrt{x^2-3}[/MATH] has no fiber when x is a positive integer"? We've been assuming you meant to say that [MATH]-\sqrt{x^2-3}= 2[/MATH] has no solution, but that isn't what you said, and it seems unlikely to be what you meant.

 

[Steven G]

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?
thanks in advance..

If two quantities are equal then if you change both signs the result will still be equal.

3=3, then -3 = -3
-9 = -9, then 9 = 9
If -x = 9, then x=-9
If -x=-2, then x=2
If x=-4, then -x = 4
If -sqrt(x^2_3) = 2, then sqrt(x^2-3)= -2. Just change the sign of both sides (it is called multiplying both sides by -1)