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I just had to prove that the derivative of the formula for solving quadratic equations was as follows:
========
We know that (x + m) squared = n
where n is greater or equal to 0
has the following solutions:
x = -m +/- sqrt n
So multiply (x + m) squared to give
x sqrd +2mx + m sqrd = n
or
x sqrd + 2mx + (m sqrd - n) = 0
a(x sqrd) + bx + c = 0 can be written as
x sqrd = (b/a)x + (c/a)x = 0
Therefore
2m = b/a and m sqrd - n = c/a
Therefore
m = b/(2a) and n = m sqrd - c/a
Therefore
n = (b/(2a)) sqrd - c/a
n = (b sqrd)/(4(a sqrd)) - c/a
n = ((b sqrd) - (4ac))/(4(a sqrd))
========
Does it look right to you?
========
We know that (x + m) squared = n
where n is greater or equal to 0
has the following solutions:
x = -m +/- sqrt n
So multiply (x + m) squared to give
x sqrd +2mx + m sqrd = n
or
x sqrd + 2mx + (m sqrd - n) = 0
a(x sqrd) + bx + c = 0 can be written as
x sqrd = (b/a)x + (c/a)x = 0
Therefore
2m = b/a and m sqrd - n = c/a
Therefore
m = b/(2a) and n = m sqrd - c/a
Therefore
n = (b/(2a)) sqrd - c/a
n = (b sqrd)/(4(a sqrd)) - c/a
n = ((b sqrd) - (4ac))/(4(a sqrd))
========
Does it look right to you?
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1+1 = 3
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