Is the Quadratic Formula the Only Solution to Quadratic Equations?

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SUMMARY

The discussion centers on the validity of an alternative formula derived from the quadratic formula, specifically the expression [-b^3-2abc ± sqrt(b^6-12a^2b^2c^2-16a^3c^3)]/(2ab^2+4a^2c). Participants debate whether this formula is a legitimate solution to quadratic equations. Daniel concludes that the alternative formula is ineffective and ultimately "useless," emphasizing that it does not provide any additional value compared to the standard quadratic formula, \(\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).

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abia ubong
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while working on the quadratic formula in school ,i came across another formula,and wanted to know if its been derived since i got this from the quadratic formula.the formula is as follows
[-b^3-2abc +or- sqrt(b^6-12a^2b^2c^2-16a^3c^3)]/(2ab^2+4a^2c).
please i want to know if it works for all.
 
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Are u asking whether

\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-b^{3}-2abc\pm\sqrt{b^{6}-12a^{2}b^{2}c^{2}-16a^{3}c^{3}}}{2ab^{2}+4a^{2}c}

...?There's only one way to find out:CROSS MULTIPLY...:devil:


Daniel.
 
looks like the multiplication of both the nominator and the denominator by
b^{2}+2ac}
 
regardless of if it works or not the form on the right is highly useless.
 
that was a hard thing to say inha, how useless is it ,i just came across it and felt like letting the forum know how correct it is and u call it useless.
thae was a hard thing 2 say
 
I'm sorry not to share your disspointment,but it's nothing more than a waste of ink and paper (or server space)...


Daniel.
 

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