- #1
mathmajor2013
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Let F be a field. Suppose f(x)=ax^2+bx+c is an element of F[x] with (2a) a unit in F, and set delta=b^2-4ac. Prove the following version of the quadratic formula. Hint: for (a) and (b) use the identity 4a x f(x)=(2ax+b)^2-delta.
(a) If there is a k in F such that k^2=delta, then (-b plus or minus k)/(2a) are roots of f(x).
(b) If there is no k in F such that k^2=delta, then f(x) has no roots in F.
(c) Use the quadratic formula to factor 3x^2+2x+5 in (Z/pZ)[x] into irreducibles for primes p in {7,11,13,23}.
I have no idea how to do this one. I started with part (c), by factoring these things. Then I tried to figure out k for these, and it gave roots that were completely different than the ones I got when I factored, which has confused me more. Help please.
(a) If there is a k in F such that k^2=delta, then (-b plus or minus k)/(2a) are roots of f(x).
(b) If there is no k in F such that k^2=delta, then f(x) has no roots in F.
(c) Use the quadratic formula to factor 3x^2+2x+5 in (Z/pZ)[x] into irreducibles for primes p in {7,11,13,23}.
I have no idea how to do this one. I started with part (c), by factoring these things. Then I tried to figure out k for these, and it gave roots that were completely different than the ones I got when I factored, which has confused me more. Help please.