"non-existant" root exist?

[Vastor]

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"button error"

 

 

hey guys. as we know that

[math]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} [/math]

 

 

where the discriminant is

[math] {b^2 - 4ac} [/math]

 

and I thought discriminant is taken from the quadratic formula, which squaroot the result of discrimination. So, that's mean when discriminant's result is negative, the answer is not available. Yet, I want to explore more about the 'un-available root', I use calculator to calculate the equation that its discriminant is negative. Then, it bugging my mind somehow the x is founded!

 

[math] x^2 - 3x + 4 = 0 [/math]

[math] D = -7 < 0 [/math]

[math] x = 1.5. [/math]

 

did I miss anything here?

Edited February 4, 2012 by Vastor

[Cap'n Refsmmat]

You can easily check for yourself that:

 

[math]1.5^2 - 3(1.5) + 4 = 1.75[/math]

 

so [math]x=1.5[/math] is not a solution. Instead, it seems your calculator is telling you the real part of the complex solution, which is:

 

[math]x = \frac{1}{2}(3 \pm i \sqrt 7)[/math]

 

where [math]i = \sqrt{-1}[/math].

[Vastor]

k, thnx

[The french tourist]

You can easily check for yourself that:

 

[math]1.5^2 - 3(1.5) + 4 = 1.75[/math]

 

so [math]x=1.5[/math] is not a solution. Instead, it seems your calculator is telling you the real part of the complex solution, which is:

 

[math]x = \frac{1}{2}(3 \pm i \sqrt 7)[/math]

 

where [math]i = \sqrt{-1}[/math].

 

Well, it seems to me that [math]i^{2} = -1[/math] would be more rigorous than [math]i = \sqrt{-1}[/math]. Because if it was the case, we could also write [math]-i = \sqrt{-1}[/math] and everyone knows the function [math]x\mapsto y=\sqrt{x}[/math] is a bijection (for every x, there is one y).

 

Vastor, you will see that when [math]\Delta < 0[/math], the solutions are the complex numbers [math]Z = \frac{-b \pm i \sqrt{-\Delta}}{2a} [/math].

[DrRocket]

Well, it seems to me that [math]i^{2} = -1[/math] would be more rigorous than [math]i = \sqrt{-1}[/math].

 

It is not a matter of rigor. It is simply a matter of notation.

 

Because if it was the case, we could also write [math]-i = \sqrt{-1}[/math] and everyone knows the function [math]x\mapsto y=\sqrt{x}[/math] is a bijection (for every x, there is one y).

 

1. The function [math]x\mapsto y=\sqrt{x}[/math] for [math]x \in \mathbb R^+[/math] is well-defined only by the convention that [math]\sqrt x[/math] selects the non-negative square root.

 

 

2. The requirement that for every x there is one y simply makes the function a function . To be a bijective function one also requires that no two values of x map to the same y-value. This is true of the square root function only because of the convention above and the fact that the domain of the function and range are taken to be the non-negative real numbers.

 

3. In the more general case of negative real numbers and complex numbers the situation requirs a bit more subtle touch. The square root function and the use of exponents in general is dependent on use of the logarithm. In complex analysis, the logarithm is not a simple well-defined function, but requires that one "choose a branch of the logarithm", reflecting the fact that the complex argument is defined only modulo [math] 2 \pi [/math]. Different chooices of that branch result in different values for the square root function. In older times the logarithm was viewed as a "multi-valued function", which of course violates the modern definition of the term "function" itself. So one must deal with a branch of a function that is actually defined on an appropriate Riemann surface rather than on the complex numbers itself.

 

Vastor, you will see that when [math]\Delta < 0[/math], the solutions are the complex numbers [math]Z = \frac{-b \pm i \sqrt{-\Delta}}{2a} [/math].

 

Yes and that expression manages to accomodate all of the ambiguity that is incurred in the expression [math]\sqrt{-\Delta}[/math] since both the negative and positive square roots are included in the complete expression.