# Variation of the quadratic formula?

1. Sep 20, 2005

### 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.

2. Sep 21, 2005

### krab

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.

3. Sep 21, 2005

### arildno

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: Sep 21, 2005
4. Sep 21, 2005

### 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.

5. Sep 22, 2005

### arildno

Yes, you multiply with 1 in the form of the conjugate of the numerator.
Thus, the root expression vanishes from the numerator but reappears in the denominator.
I'm not too sure about what you mean by "rationalizing" and "backwards", though..