- #1

toltol123

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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.

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- Thread starter toltol123
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- #1

toltol123

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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

krab

Science Advisor

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[tex](-b+\sqrt{b^2-4ac})(-b-\sqrt{b^2-4ac})=4ac[/tex]

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

arildno

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Alternatively, multiply with 1 in a smart manner:

[tex]\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}}[/tex]

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.

[tex]\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}}[/tex]

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:

- #4

toltol123

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- #5

arildno

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Yes, you multiply with 1 in the form of the conjugate of the numerator.toltol123 said:

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..

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