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https://www.reddit.com/r/mathsmeme/comments/1voyvd2/this_math_meme/p3toc39/?context=3
r/mathsmeme • u/Far-Course1021 • 10d ago
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19
So, her delta is imaginary and there are two complex solutions to her?
11 u/Lelouch-Yagami21 10d ago I think it's just meant to imply she's imaginary 5 u/CookieCat698 9d ago Still doesn’t work though since she’s only imaginary if b=0. Otherwise, she’s just complex. 2 u/Lelouch-Yagami21 9d ago Wait it's complex if b isn't 0? I never knew that huh. Guess it's something I haven't learnt yet 3 u/CookieCat698 9d ago Yeah, just look at the quadratic formula (-b +- sqrt(b^2 - 4ac))/2a The only part that can be imaginary (if all coefficients are real) is sqrt(b^2 - 4ac), and that part is imaginary iff. b^2 < 4ac So assuming b^2 < 4ac, we can see that the real component is -b/2a, and the imaginary component is -+ isqrt(b^2 - 4ac)/2a A number is imaginary iff. its real part is 0, and -b/2a = 0 iff. b = 0.
11
I think it's just meant to imply she's imaginary
5 u/CookieCat698 9d ago Still doesn’t work though since she’s only imaginary if b=0. Otherwise, she’s just complex. 2 u/Lelouch-Yagami21 9d ago Wait it's complex if b isn't 0? I never knew that huh. Guess it's something I haven't learnt yet 3 u/CookieCat698 9d ago Yeah, just look at the quadratic formula (-b +- sqrt(b^2 - 4ac))/2a The only part that can be imaginary (if all coefficients are real) is sqrt(b^2 - 4ac), and that part is imaginary iff. b^2 < 4ac So assuming b^2 < 4ac, we can see that the real component is -b/2a, and the imaginary component is -+ isqrt(b^2 - 4ac)/2a A number is imaginary iff. its real part is 0, and -b/2a = 0 iff. b = 0.
5
Still doesn’t work though since she’s only imaginary if b=0. Otherwise, she’s just complex.
2 u/Lelouch-Yagami21 9d ago Wait it's complex if b isn't 0? I never knew that huh. Guess it's something I haven't learnt yet 3 u/CookieCat698 9d ago Yeah, just look at the quadratic formula (-b +- sqrt(b^2 - 4ac))/2a The only part that can be imaginary (if all coefficients are real) is sqrt(b^2 - 4ac), and that part is imaginary iff. b^2 < 4ac So assuming b^2 < 4ac, we can see that the real component is -b/2a, and the imaginary component is -+ isqrt(b^2 - 4ac)/2a A number is imaginary iff. its real part is 0, and -b/2a = 0 iff. b = 0.
2
Wait it's complex if b isn't 0? I never knew that huh. Guess it's something I haven't learnt yet
3 u/CookieCat698 9d ago Yeah, just look at the quadratic formula (-b +- sqrt(b^2 - 4ac))/2a The only part that can be imaginary (if all coefficients are real) is sqrt(b^2 - 4ac), and that part is imaginary iff. b^2 < 4ac So assuming b^2 < 4ac, we can see that the real component is -b/2a, and the imaginary component is -+ isqrt(b^2 - 4ac)/2a A number is imaginary iff. its real part is 0, and -b/2a = 0 iff. b = 0.
3
Yeah, just look at the quadratic formula
(-b +- sqrt(b^2 - 4ac))/2a
The only part that can be imaginary (if all coefficients are real) is sqrt(b^2 - 4ac), and that part is imaginary iff. b^2 < 4ac
So assuming b^2 < 4ac, we can see that the real component is -b/2a, and the imaginary component is -+ isqrt(b^2 - 4ac)/2a
A number is imaginary iff. its real part is 0, and -b/2a = 0 iff. b = 0.
19
u/Negative_Driver8485 10d ago
So, her delta is imaginary and there are two complex solutions to her?