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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?
Page:
I use the following technique to derive the quadratic formula:
1) Write out the final formula
2) Rearrange it step by step until it's a quadratic equation
3) Write out all the steps you have just done, in reverse order.
An there you go, you've derived the quadratic formula from a simple ax^2 + bx + c
1) Write out the final formula
2) Rearrange it step by step until it's a quadratic equation
3) Write out all the steps you have just done, in reverse order.
An there you go, you've derived the quadratic formula from a simple ax^2 + bx + c
((b sqrd) - (4ac))/(4(a sqrd))
look at that, see the problem u need to times everything out side the bracket with everything in so the sue of the brackets around the 4 is a complete wast of time the divide makes an automatic braket anyway.
look at that, see the problem u need to times everything out side the bracket with everything in so the sue of the brackets around the 4 is a complete wast of time the divide makes an automatic braket anyway.
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?
Page: