YAY!!!! ICEE LAND!!!!!!
- Locked by Carri04 on May 24, '20 11:39pmReason: To preserve the thread, spam free.
Thread Topic: YAY!!!! ICEE LAND!!!!!!
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Thank you. :P
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You're welcome. =]
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So what's up?
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Watching Johnny Bravo on Boomerang. XP You?
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*giggles*
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I'm pretty sure 2 + 2 = 93....
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Or maybe it equals 3569844311790.... :\
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No 2+2=4. ._.
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I know. XP
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Yeah cuz I said it. ;P
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Well everyone should know that y=ax^2+bx+c at least. Right? =3
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Actually I think they should know: If ax*x+bx+c=0 then what is X? (Yes I know the answer to that. x3)
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Maybe.......... x=
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Force equals mass times acceleration XP
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No ax^2+bx+c= 0.....(1), as x*x = x^2. This is a second degree equation and is also called a quadratic equation which will be solved by writing the left as diffrence of an unknown square and a known square equal to zero:
LHS of (1) :
ax^2+bx+c = a[x^2+(b/a)x]+c
=a(x^2+(b/a)x+(b/(2a))^2] - b^2/(4a)+c. Here (b^2/(4a) is added and subtracted.
=a[x+(b/(2a))]^2 - (b^2-4ac)/4a .
=a{[x+b/(2a)]^2 - (b^2-4ac)/(2a)^2}
So the equation (1) now could be rewritten like:
a{[x+b/(2a)]^2 - (b^2-4ac)/(2a)^2} = 0. Or bt dividing by a which is not equal to zero, we get:
[x+b/(2a)]^2 - (b^2-4ac)/(2a)^2 = 0. Or
[x+b/(2a)]^2 = (b^-4ac)/(2a)^2. Taking the square root,
x+b/(2a) = +sqrt(b^2-4ac)/2a . Or x+b/(2a) = -sqrt(b^2-4ac)/2a. So we get 2 solutions from the se two results.
x1 = [-b+sqrt(b^2-4ac)]/(2a) Or
x2 = [-b-sqrt(b^2-4ac)]/(2a)
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