Is the Quadratic Formula the Only Solution to Quadratic Equations?

  • Context: High School 
  • Thread starter Thread starter abia ubong
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Discussion Overview

The discussion centers around the validity and utility of an alternative formula derived from the quadratic formula for solving quadratic equations. Participants explore whether this new formula is applicable to all quadratic equations and its overall usefulness compared to the standard quadratic formula.

Discussion Character

  • Debate/contested, Technical explanation

Main Points Raised

  • One participant presents an alternative formula derived from the quadratic formula and questions its validity for all quadratic equations.
  • Another participant suggests verifying the equivalence of the two formulas through cross-multiplication.
  • A different participant comments on the multiplication of terms in the alternative formula, indicating a potential simplification.
  • Some participants express skepticism about the usefulness of the alternative formula, with one describing it as "highly useless" and another agreeing that it may be a waste of resources.
  • Another participant defends the alternative formula, expressing disappointment at its dismissal and asserting its correctness.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the validity or usefulness of the alternative formula. There are competing views regarding its applicability and relevance.

Contextual Notes

Participants have not fully explored the assumptions or conditions under which the alternative formula might hold true, nor have they resolved the mathematical steps necessary to establish its equivalence to the standard quadratic formula.

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.
 
Mathematics news on Phys.org
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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