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|#3 - S. Negative 6. Moses. [+] (19 new replies)||12/16/2014 on Going to class drunk||+120|
#8 - lolollo (12/16/2014) [-]
Teacher: "What's the square root of 49?"
Guy in class: "7"
Everyone: "The fuck?"
Teacher: "It's positive and negative 7!"
Everyone's face when the teacher then rants and raves for 10 minutes about how we have to study up on our square roots.
#29 - anonymous (12/17/2014) [-]
ok faggots, time to learn math:
let's assume sqrt(x) = ±y, therefore, sqrt(x) = -sqrt(x), as ±y = -(±y) = ±y (the negative sign just reverses any sign, but since it's a ± sign, it is the same reversed). if we have sqrt(x) = -sqrt(x), then sqrt(x) + sqrt(x) = 0 (add sqrt(x) to both sides)
now, with the assumption that sqrt(x) = ±y, then with this we get: ±y + ±y = 0, or ±y ± y = 0
now, analyzing all the possible signs you could use with this equation: y + y = 2y, -y + y = 0, y - y = 0, -y - y = -2y
so, we get that sqrt(x) = -sqrt(x) only 50% of the time, so from this we determine that sqrt(x) =/= -sqrt(x), and with that, we get that sqrt(x) =/= ±y, thus disproving ±7 being the root of 49
principal root: sqrt(x) will always return the positive root of x, and -sqrt(x) will always return the negative root of x
tl:dr i'm siding with civilizedwasteland on this one inb4 redthumbed to oblivion
#31 - anonymous (12/17/2014) [-]
now, you may have some arguments
1. "you substitute + or - for the ± when adding ±y and ±y"
yes you do, in the quadratic formula both (-b+sqrt(b^2-4ac))/2a and (-b-sqrt(b^2-4ac))/2a are solutions, so you do have to use both possibilities for the signs.
2. "the signs you use when adding ±y and ±y must be opposites of each other"
?, why would that be the case? if sqrt(x) = ±y, then it shouldn't matter whether they're opposites signs or not because it should remain true in all cases.
3. "-(±y) =/= ±y"
add ±5 to 5, then try subtracting ±5 from 5, same answer(s)
#26 - terminallyunique (12/17/2014) [-]
>refers to imaginary or complex numbers
Nigga what the hell do you think complex numbers are ?
You CAN find the square root of a negative number; thats the whole concept of imaginary and complex numbers. Use "i" to represent the root of -1; therefore i^2 = -1
By extension root -49 for example, is 7i.
But that's not even the point. Every number has a square root. If you square a negative value you get a positive value. That's why in saying a value for square root answers, unless otherwise stated, the answer is always +/- the actual numerical root value.
Source : Current Pure and Applied Mathematics student. (I swear I'm 18)
#19 - anonymous (12/17/2014) [-]
Technically sqrt(49) can only be 7 because it is the principal square root, to get sqrt(49) to be equal to -7 you would have to say -sqrt(49). the reason x^2=49 => x=+/- 7 is because you can input either -7 or 7 into the equation to get 49, with just the sqrt function you only use the positive answer as you can only return one solution and it is also accepted to only use the positive root.
sqrt(x) =/= -sqrt(x)
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