Is the Quadratic Formula the Only Solution to Quadratic Equations?

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The discussion centers on an alternative formula derived from the quadratic formula, questioning its validity for solving quadratic equations. Participants debate whether this new formula is equivalent to the standard quadratic formula. One contributor suggests testing the equivalence through cross multiplication, while another criticizes the alternative formula as impractical and unnecessary. The consensus leans towards the view that the new formula is not useful and may not provide any additional benefits. Ultimately, the conversation emphasizes the established effectiveness of the traditional quadratic formula over the proposed alternative.
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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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