Let $K$ be any subfield of $\mathbb{C}$ and let $m(t)$ be a quadratic polynomial over $K$. show that all zeros of $m(t)$ lie in an extension $K(\alpha)$ of $K$ where $\alpha^2=k\in K$. Thus allowing square roots $\sqrt k$ enables us to solve all quadratic equations over $K$.
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Obviously the quadratic formula works. But you might wonder why it works. Here's a nice way of understanding why. Suppose more generally that $\rm\,K\,$ is a field of characteristic $\ne 2\,$ (so that we can divide by $\,2\,$ below) and suppose we have a calculator that can compute the field operations of $\rm\,K\,$ as well as square roots of elements in $\rm\,K\,$. Then given the sum $\rm\, b = r+s\,$ and product $\rm\,c = r\,s\,$ of any two elements $\rm\,r,s\in K,\,$ we can solve for $\rm\,r,s\,$ on our calculator. To do so it suffices to find their difference $\rm\,d = r\!-\!s\,$ since $\rm\ r = (r\!+\!s + r\!-\!s)/2 = (b\!+\!d)/2\ $ and $\rm\ s = (r\!+\!s)-r = b\!-\!r\,.\,$
But how can we find their difference $\rm\,d\,$? By exploiting innate symmetry: $\rm\,d = r\!-\!s\, $ is almost symmetric, i.e. it only changes sign when we swap $\rm\,r\,$ and $\rm\,s\,.\,$ We can eliminate this sign change by squaring it. Therefore, since $\rm\,d^2\,$ is a symmetric polynomial in $\rm\,(r,s),\,$ by the fundamental theorem of symmetric polynomials, it can be expressed as a polynomial in the elementary symmetric polynomials on $\rm\,(r,s),\,$ namely $\rm\ r+s,\ \ r\ s\,.\,$ While we could use Gauss's algorithm to find this polynomial, here it is rather obvious, namely $\rm\ (r+s)^2 - (r-s)^2 = \ 4\,r\,s,\ $ i.e. $\rm\ b^2 -d^2 = 4\,c,\ $ so $\rm\,d^2 = b^2-4\,c\,.\,$ Thus we can compute $\rm\,d\,$ by taking the square root of $\rm\,b^2-4\,c,\,$ and this yields $\rm\,r,s\,$ via the equations above. So we can recover any two numbers from their sum and product - by way of a symmetry-derived equation relating the sum, product and difference.
This is the essence of the genesis of the quadratic formula. Indeed, by Vieta, $\rm\,r,s\,$ are roots of $\rm\, (x-r)\ (x-s)\, =\, x^2 - b\ x + c\, $ with sum $\rm\,b,\,$ product $\rm\,c\,$ and discriminant $\rm\ b^2 - 4c\, $ so the grade-school quadratic formula amounts to the same formula derived above using symmetry. The advantage of the above viewpoint is that it serves to better reveal the innate symmetries -- something that will become much clearer when one studies Galois theory. For a taste see the section on Lagrange resolvents in the Wikipedia quadratic formula page.
Bill Dubuque
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See http://en.wikipedia.org/wiki/Viète's_formulas . – Bruno Stonek Mar 30 '11 at 06:12
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the above is of course true, but I also recall the existence of a 'purely algebraic' proof (in terms of field extensions etc.) for this fact. If anyone has any ideas, they'd be very welcome. – Gerben Mar 30 '11 at 08:18
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@Bruno: @Gerben: It's not simply Vieta's formulas. The point of my post is to encourage the reader to revisit the grade school proof of the quadratic formula from the more advanced viewpoint of Galois theory. – Bill Dubuque Mar 30 '11 at 12:36
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2@Bill Dubuque: There are a number of Babylonian "word problems" that are equivalent to solving $x+y=a$, $xy=b$. The solutions (implicitly) use the identity $(x-y)^2+4xy=(x+y)^2$. – André Nicolas Jul 03 '11 at 21:19
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I don't see a section on Lagrange resolvents in the Wikipedia quadratic equation page – J. W. Tanner Oct 01 '20 at 13:41
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1@J.W.T It was moved from the original page on qudratic equations to a page on the quadratic formula - see the updated link in the answer/ – Bill Dubuque Oct 01 '20 at 20:24
