Is the Quadratic Formula the Only Solution to Quadratic Equations?

  • Thread starter abia ubong
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In summary, during a discussion about the quadratic formula, a forum member brought up a different formula and asked if it had been derived. The formula involves a fraction with a square root in the numerator and a polynomial expression in the denominator. Another member responded by explaining that this formula is equivalent to the quadratic formula and is not useful. However, the original member still wanted to know if it works for all cases and suggested cross multiplying to find out.
  • #1
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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  • #2
Are u asking whether

[tex] \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} [/tex]

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


Daniel.
 
  • #3
looks like the multiplication of both the nominator and the denominator by
[tex] b^{2}+2ac} [/tex]
 
  • #4
regardless of if it works or not the form on the right is highly useless.
 
  • #5
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
 
  • #6
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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