Quadratic Equation Error For the floating point system $(B, t, L, U) = (10,8,-50,50)$ and for the quadratic equation:
$ax^2 + bx + c$, I need to show error that arises in various cases and how to fix those. 
$$a=10^{-30}$$
$$b= -10^{30}$$
$$c=10^{30}$$
I think there is cancellation error when plugging into quadratic formula. Can anyone help me verify that if multiply quadratic formula by conjugate, then I'll get $\frac{2c}{-b \pm \sqrt{b^2-4ac}}$ which will take away the error for the case that is subject to error?
Thanks!
 A: OK, just for fun, let's try a different method:
$$
ax^2 + bx + c = 0
$$
$$
a + b\frac 1 x + c\frac{1}{x^2} = 0
$$
$$
a + bu + cu^2 = 0
$$
So $u=1/x$, and now we have a quadratic equation in $u$.  Solve that by the usual formula (but we've interchanged the customary roles of $a$ and $c$, so do that in the formula as well):
$$
u = \frac{-b\pm\sqrt{b^2 - 4ca}}{2c}.
$$
$$
x = \frac{2c}{-b\pm\sqrt{b^2-4ca}}.
$$
A: The basic idea is that if |$4ac| \ll b^2$, the square root is very close to $|b|$.  Depending on the sign of $b$, one combination or the other will involve subtracting two nearly equal quantities.
To be specific, let us assume $b \gt 0$.  Then one root is $\frac {-b+\sqrt{b^2-4ac}}{2a}$ and is the one subject to cancellation.  If we multiply by the congugate:$$\frac {-b+\sqrt{b^2-4ac}}{2a}\frac {b+\sqrt{b^2-4ac}}{b+\sqrt{b^2-4ac}}=\frac{b^2-4ac-b^2}{2ab+2a\sqrt{b^2-4ac}}=\frac{-2c}{b+\sqrt{b^2-4ac}}$$ and the cancellation has disappeared.  The case $b \lt 0$ is similar except you worry about the one with the minus sign.
