quadratic equation - discrimination

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- Dec 18th 2007, 10:40 PMafeasfaerw23231233quadratic equation - discrimination
quadratic equation - discrimination

- Dec 19th 2007, 12:03 AMmr fantastic
Your expression for the discriminant can be re-arranged into

$\displaystyle D = 4(a^2 - a(b+c) + (b^2 - bc + c^2))$.

For unequal roots (ie. two distinct roots) you want $\displaystyle a^2 - a(b+c) + (b^2 - bc + c^2) \neq 0$.

NB (an acronym for the latin phrase*nota bene*which translates as*note well*, that is,*what follows is very important so pay close attention, chum*): We don't give a fetid dingos kidney whether the roots are real or not, as long as they're distinct. So the requirement is $\displaystyle D \neq 0$, NOT $\displaystyle D > 0$.

Let's assume $\displaystyle a^2 - a(b+c) + (b^2 - bc + c^2) = 0$. Since a is real, then (IF the assumption is true) you can solve*this*quadratic to find real values of a. Therefore (IF the assumption is true) the discriminant of*this*quadratic is greater than zero.

From here, I'll let you fill in the steps that lead to $\displaystyle (b - c)^2 < 0$, an impossibility for real values of b and c.

Hence the assumption $\displaystyle a^2 - a(b+c) + (b^2 - bc + c^2) = 0$ is false. So $\displaystyle a^2 - a(b+c) + (b^2 - bc + c^2) \neq 0$. Q.E.D. - Dec 19th 2007, 04:29 AMJaneBennet
You can also write

$\displaystyle D\ =\ (2a-b-c)^2+3(b-c)^2$

Since $\displaystyle b\neq c,\ 3(b-c)^2\ >\ 0$.

Hence*D*is (strictly) positive, so the equation has real and unequal roots.