# Variation of the quadratic formula?

toltol123

There's another way of writing the quadratic formula...but how do I derive the 2nd formula using the 1st one? I've tried for hours and I can't get it. I would really appreciate any help.

Hint:
$$(-b+\sqrt{b^2-4ac})(-b-\sqrt{b^2-4ac})=4ac$$
Take the two different forms, set them equal to each other and simplify. But you have to know that the lower sign of the plus/minus in the one form corresponds to the upper sign in the other form.

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Alternatively, multiply with 1 in a smart manner:
$$\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}*1=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}*\frac{-b\mp\sqrt{b^{2}-4ac}}{-b\mp\sqrt{b^{2}-4ac}}=\frac{b^{2}-(b^{2}-4ac)}{2a(-b\mp\sqrt{b^{2}-4ac})}=\frac{2c}{-b\mp\sqrt{b^{2}-4ac}}$$

The alternate version is often used if the numerator in the standard version becomes a difference between almost equal numbers.
I.e, we may avoid loss of significant digits by using the alternate version.

Last edited:
toltol123
oh, so in a way I'm rationalizing the equation and just multiplying by the conjugate? except it's backwards, because the radical is moved to the bottom.