@Guardian Talon
Variables in quadratic formulas can have multiple answers, actually. That’s kind of the entire point of them. Both -2 and -3 are possible solutions though obviously it can’t be both at the same time, no. But;
@Background Pony #89D9
Again, no need to go full quadratic when you’re dealing with basic factoring.
6 is a product of 2 and 3, which add up to 5. Its about as basic as you can get.
(x+2)(x+3)=0
x=-2 and x=-3
Well, she’s almost right - the 0 is a lie. x = a bunch of stuff = 0. x = 0 = a bunch of stuff. but the bunch of stuff can’t be zero because x is already zero, and that makes the bunch of stuff 6. Therefore six is zero. So the damage has to be done to the six, to knock the top bit off of it.
I see. Thanks.
I kinda upset my math teachers in high school so I wasn’t aloud to take the advanced stuff.
Variables in quadratic formulas can have multiple answers, actually. That’s kind of the entire point of them. Both -2 and -3 are possible solutions though obviously it can’t be both at the same time, no. But;
(-3+2)(-3+3) -> (-1)(0) = 0
(-2+2)(-2+3) -> (0)(1) = 0
Or unfactored, since you claim it can’t be -3
(-3) 2 + 5(-3) + 6
9 - 15 + 6
15 - 15 = 0
QED.
X can’t =-2 and -3 at the same time that would be saying -2=-3
The answer is -2
(-2)* (-2)+ 5*(-2)+ 6 = 4+(-10)+6 = 0
Again, no need to go full quadratic when you’re dealing with basic factoring.
6 is a product of 2 and 3, which add up to 5. Its about as basic as you can get.
(x+2)(x+3)=0
x=-2 and x=-3
Originally “Solve for x:” was there, but Pinkamena incorrectly copied it as “x=”
x1= -2
x2= -3
But Pinkamena’s method is much more fun. :D
HELP
shuffles incoherently
fixes tag