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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 that you have a calculator that can compute the field operations of $\rm\:K\:$ as well as square roots of elements in $\rm\:K\:$. Then I claim that given the sum $\rm\: b = r+s\:$ and product $\rm\:c = r\:s\:$ of any two elements $\rm\:r,s\in K\:,\:$ you can solve for $\rm\:r,s\:$ on your 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 $\rm\:d\:$? With a little help from symmetry. Note that $\rm\:d = r-s\ $ is almost symmetric, i.e. it merely 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. Therefore 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 quadratic formula. Indeed, $\rm\:r,s\:$ are roots of $\rm\: (x-r)\ (x-s)\: =\: x^2 - b\ x + c\ $ with the discriminant $\rm\ b^2 - 4\:c\:,\ $ so the grade-school quadratic formula amounts to the same formula derived above. 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 equation page.

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    @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 $(x-y)^2+4xy=(x+y)^2$.2011-07-03
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Hint: Use the quadratic formula.

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    Thanks!!!!! no se porque no me hhabia dado cuenta!!!!!2011-03-30