an example in my text says that if the values for a, b and c are 2π, 2πh and -S respectively

with those values plugged into the quadratic equation you get

-2πh ± πi^2 h^2 - 4(2π)(-S)
--------------------------------------------------------------
4
π

up to that point i follow what is going on but then the text reduces the formula to

-2
πh ±2 π^2 h^2 - 2πS
------------------------------------------------------
4
π

what I don't understand is how -4(2
π)(-S) reduces to 2πS in the above equation.

from there we divide everything by 2 and get

-
πh ± πi^2 h^2 + 2πiS
---------------------------------------------
2
π

2. ## Re: quadratic equation reduction

Originally Posted by kingsolomonsgrave
an example in my text says that if the values for a, b and c are 2π, 2πh and -S respectively

with those values plugged into the quadratic equation you get

-2πh ± πi^2 h^2 - 4(2π)(-S)
--------------------------------------------------------------
4
π

up to that point i follow what is going on but then the text reduces the formula to

Unfortunately, it is this part, that you say you follow, that is wrong.
$a= 2\pi$, $b= 2\pi h$, and $c= -S$ so the "discriminant" inside the square root should be
$b^2- 4ac= (2\pi h)^2- 4(2\pi)(-S)= 4\pi^2h^2+ 8\pi S$
You are missing the "4" in first term.

Once you have that, you can factor out a "4" so that $\sqrt{4\pi^2h^2+ 8\pi S}= 2\sqrt{\pi^2h^2+ 2\pi h}$.

-2
πh ±2 π^2 h^2 - 2πS
------------------------------------------------------
4
π

what I don't understand is how -4(2
π)(-S) reduces to 2πS in the above equation.

from there we divide everything by 2 and get

-
πh ± πi^2 h^2 + 2πiS
---------------------------------------------
2
π