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

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SUMMARY

The discussion centers on the correct method for solving a quadratic equation involving absolute values, as illustrated in two images (pic 1 and pic 2). Participants agree that both methods yield the same solutions, but there is contention over the validity of the approaches. The consensus is that the method used in pic 2, which simplifies the absolute value directly, is more comprehensive as it accounts for all possible cases. Misunderstandings arise from the interpretation of square roots and absolute values, emphasizing the importance of clarity in mathematical conventions.

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  • Knowledge of absolute value properties and their implications in equations.
  • Familiarity with square root conventions and their mathematical significance.
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  • #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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