Maths Question

@Disch this question does not delve into imaginary numbers, as far as I know
@L B Excuse my ignorance, but what does the graph tell me?

sqrt(x - 1) = -3
x - 1 = (-3)^2
x - 1 = 9
x = 10?
Ignore all of that.

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((x-1)^0.5) + 3 = 0 iff sqrt(x-1) = -3 iff sqrt(y) = -3.
By definition of sqrt: for all real values of q, sqrt(q) is non-negative.
Therefore, there exists no real x that meets the condition.

Disch (13742)

...which is why I thought x had to be imaginary. But apparently there's not even an imaginary solution?

Imaginary numbers are when you have a square root of a negative number, not when the square root of a number equals a negative number.

sqrt(-4) = 2i

I was taught that sqrt(9) was equal to +/- (read: plus minus) 3

Hmm this is interesting. Maybe helios can elaborate here.

BigBlackSheep (35)

By definition of sqrt: for all real values of q, sqrt(q) is non-negative.

Otherwise f(x) = x^0.5 wouldn't be a function.

I read an interesting thing here: http://www.xamuel.com/plus-or-minus-square-roots/

For example, 4 has two square roots, 2 and -2. This is because 2²=4 and (-2)²=4. But the square root of 4 is 2 (never -2).

And was wondering why? (The later explanation fails to convince me)

I was taught that sqrt(9) was equal to +/- (read: plus minus) 3

Definition of sqrt: sqrt is a function such that
1. If x>=0, sqrt(x) is a non-negative number such that sqrt(x)^2 = x.
2. If x<0, sqrt(x) is undefined.

You could define a multifunction f:C->P(C) such that: for all y in f(x), y^2 = x <edit>and there exist no other complex numbers that meet this condition</edit> (f(x) is a set). But then obviously f != sqrt.

...which is why I thought x had to be imaginary. But apparently there's not even an imaginary solution?

Imaginary numbers are when you have a square root of a negative number, not when the square root of a number equals a negative number.

(-sqrt(3)i)^2 = -3 (i^2) = -3 (-1) = 3

EDIT:

And was wondering why? (The later explanation fails to convince me)

Like that page says, there's a ~~definition~~ distinction between a square root, which is just any number whose square is equal to the given number, and the square root, which is a mathematical function with a precise definition. Because the square root of x (or sqrt(x), or x^0.5) is a function, it is one number and one number alone. There's no technical reason why the image of sqrt is [0;+inf.). It's just so by convention. Whatever convention you choose, exactly one of these equations has no solution (provided that both sqrt follow the same convention):
* sqrt(x)=3
* sqrt(x)=-3

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This is weird. Such an elementary mathematics topic, and I'd never really thought of this. I understand a function can only have one solution per input. But then how does a square root of a positive number have 2 solutions? Why does the square root have to be a function?

@helios I think I get it, the reason my first mentioned equation is false, is simply due to my prior lack of understanding of the definition of the sqrt function. thank you :)

EDIT: which is sqrt(x)=b; b>=0

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See my edit.

BigBlackSheep (35)

I'm just curious ResidentBiscuit, I've seen a lot of your posts suggesting you're in your 3rd/4th (?) year of university for computer science. Does your school not require you to take many math courses? Or is sqrt just something you never really thought about and always just assumed that the positive value was the "right" one? (This is not meant to be offensive, I'm just curious)

LB (13399)

+(4^½) = +(2)
-(4^½) = -(2)
±(4^½) = ±(2)
∓(4^½) = ∓(2)

Think about it, what if +4^½ = ±2?
2² = 4 = 2 * 4^½ = 2 * ±2
4 = ±4 -> 4 = 4 and 4 = -4

An even root does not imply ± nor ∓, you have to specify that with the root.

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