Quadratic equation: Which way is correct? pic1 or pic2?

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The discussion centers on the correct method for solving a quadratic equation involving absolute values, comparing two approaches depicted in images. One method (pic 1) adheres to mathematical properties regarding absolute values, while the other (pic 2) simplifies the equation by removing absolute values and directly setting it equal to positive and negative values. Both methods yield the same solutions, but concerns arise about the validity of the second method if conditions do not match. The consensus is that while both methods can arrive at correct answers, the professor's approach is more comprehensive as it considers all possible cases. Ultimately, clarity in mathematical conventions and problem presentation is emphasized to avoid confusion.
  • #31
As the convention
\sqrt{}
denotes nonnegative square root , e.g. 2 out of 2,-2 which are square roots of 4.

That is OK but I am puzzled in using this symbol for negative or complex numbers.
For an example
\sqrt{-1}=i
Why not -i ? What is the convention ? I suppose it is "nonnegative on pure imaginary axis". Is it right?
Square roots of i are ##e^{\pi/4\ i},e^{5\pi/4\ i}##. Which is ##\sqrt{i}## ?
 
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  • #32
anuttarasammyak said:
That is OK but I am puzzled in using this symbol for negative or complex numbers.
For an example
\sqrt{-1}=i
Why not -i ? What is the convention ? I suppose it is "nonnegative on pure imaginary axis". Is it right?
Square roots of i are ##e^{\pi/4\ i},e^{5\pi/4\ i}##. Which is ##\sqrt{i}## ?
"Nonnegative on pure imaginary axis" seems to be the convention. For your second question, several web sites I looked at (search on "principal square root of a complex number") define the principal square root of a complex number as the root with a positive imaginary part. It's important to note that the imaginary part is the coefficient (a real number) of i. This would make ##e^{\pi/4 i}## the principal square root of i.
 
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  • #33
Thank you so much @Mark44 .

So we may summarize the convention of ##\sqrt{}## that we choose from square roots
#1 nonnegative one on imaginary axis
If imaginary part are zero,
#2 nonnegative one on real axis.
Or in a word it has phase angle of ##\phi[0,\pi)##.

\sqrt{e^{i5\pi/4}}=e^{i5\pi/8}
whose real part is negative and imaginary part is positive. Imaginary axis prevails.
 
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