what is the degree of the following extension over Q -- Q(sqrt(3+2*sqrt(2)))?
115194
3 + 2sqrt(2) = (1 + sqrt(2))^2 so sqrt(3+2sqrt(2))=1+sqrt(2).\n\nBugzpodder, (x^2-3)^2 - 8 = x^4 - 6 x^2 + 1 is reducible.  Namely, it is x^4 - 2 x^2 + 1 - 4 x^2 = (x^2-1)^2 - (2x)^2\n   = (x^2 - 2x - 1)(x^2 + 2x - 1)
1048436
As a^2 is contained in F(a), clearly F(a^2) is contained in F(a). So we need to show the reverse inclusion.\n\nWe have the chain of fields F < F(a^2) <= F(a).\n\n(where I use <= and < for subset and proper subset containment) \n\nand hence [F(a) : F] = [F(a) : F(a^2)][F(a^2) : F]. Set n = [F(a) : F] and m = [F(a^2) : F]. The preceeding equation implies that m divides n. It will suffice to that m = n. Let f(X) = X^m + c_{m-1}X^{m-1} + ... + c_1X + c_0 denote the minimal polynomial of a^2 over F. Then f(X^2) = X^{2m} + c_{m-1}X^{2(m-1)} + ... + c_1X^2 + c_0 has a as a root. This implies that the minimal polynomial of a has degree less than or equal to 2m. So we get m divides n and n <= 2m. The only ways for this to occur are when m=n or when 2m = n. But n is odd, so n does not equal 2m. Hence m = n. This completes the proof.
816170
You have done some mistake while copying the problem.Please check it up.It may be 36l^2-36lq+9q^2 or something else.However I am doing ipresuming it to be 36l^2-36lq+9q^2\n\ngiven exp.=9(4l^2-4lq+q^2)\n=9{2l)^2-2.2l.q+(q)^2}\n=9(2l-q)^2\nThe answer may also be written as {3(2l-q)}^2 or (6l-3q)^2
864310
If K is a Galois extension field of F, the you know that the degree of the Galois group is [K:F]. Thus, G is a cyclic group of degree [K:F].\n\nBy the fundamental theorem of Galois theory, every subgroup of G corresponds to a field intermediate between F and K and vice versa. If H is a subgroup, then the corresponding subfield is the stabilizer of H. So the question boils down to what the subgroups of a cyclic group are. \n\nBut subgroups of cyclic groups are cyclic with order dividing the order of G. In fact, there is exactly one subgroup of order k for each k dividing the order of G. Furthermore, the index of a subgroup in G is the degree of the extension from F to the subfield corresponding to that subgroup.
756805
x<2   (-infinity,2)\nx>3   (3, infinity)\nx<2 see first interval\nx>-2 (-2,infinity)
1007589
1. b^2-4ac>0 : b>2sqrt(ac)\n2. b^2-4ac=0 : b=2sqrt(ac)\n3. b^2-4ac<0 : b<2sqrt(ac)\n\nThe rest should be a piece of cake.
296703
-3 + sqrt(2) or -(3 - sqrt(2))\nthats like saying\n\n-1x, 0x, 1x\n\nthe answer is problem found like this\n\n2x, 3x, 4x\n\nso i believe the answer is \n\n6 - 2sqrt(2), 9 - 3sqrt(2), 12 - 4sqrt(2)
245176
u*(u-2)+v*(u-2)\n(u+v)*(u-2)
837237
Hi\nI think you're referring to one of my answers.\nhttp://answers.yahoo.com/question/index;_ylt=ApZ1XohHNUdL5xkBg8bXQEfsy6IX?qid=20061124162535AAURdJ5\n\nHy is correct, this was done in order to eliminate the x^3 term in the quartic equation.  It made the subsequent solution much easier to handle, since the actual values of a,b,c,d or p,q,and r didn't matter:  the solution wasn't looking for an answer to a specific quartic but for a generalization about quartics in general.\n\nConverting a quartic into reduced form is also done when finding the roots of any quartic equation by an old method (hardly used anymore since computers).  After placed into reduced form, the 'cubic resolvant' is formed:\n\nz^3 + 2pz^2 + (p^2 - 4r)z - q^2 = 0\n\nFinding the three roots (z1,z2,z3) of this equation (using Cardano's or other methods), will yield the four roots of the reduced form quartic:\n\nx1 = (√z1 + √z2 + √z3) / 2\nx2 = (√z1 - √z2 - √z3) / 2\nx3 = (-√z1 + √z2 - √z3) / 2\nx4 = (-√z1 - √z2 + √z3) / 2
46351
Sy   = |x| - x^2 + 2   \n     = (x^2/2) - (X^3/3) + 2X
851139
If you had a normal polynomial like x^3 + 2x + 1, the degree would be 3. Thats because it is the higher exponent that occurs over all terms.\nWhen you have multiple variables, to find the degree you act as if they were all the same - so x^2 y has degree 3, 3x^2 has degree 2, z has degree 1, so overall the degree is 3.\nIn the second case, x^3y contributes the most (4), so it has degree 4.\n\nedit - just responding to the guy who posted after me, since I don't want you to get confused.\nQuoting his source:\nDegree: The highest degree of any term in the polynomial.\nDegree of term: For a term with one variable, the degree is the variable's exponent. With more than one variable, the degree is the sum of the exponents of the variables.\n\nThus my answer was correct :)
598233
just a little correction to taylor's answer.\nsqrt(x-2)=1, square it on both sides \n=> x-2 => square(1)=1\n=> x=3
874890
the degree of the variable is it's power like : A to the power 3\nthen A is a third degree variable....\n                                                                                         bye :)
898112
The math is the same as the last one.\n\nDegree (F) = (Degree (C) * (9/5) + 32\nTherefore 44 C = 111.2 F
720929
Let me just work through the steps to solve the problem:\n\nsqrt(2x^2-1)=x   <---Original Equation\n2x^2-1=x         <---Square both sides\n2x^2-x-1=0       <---Subtract x\n\nNow, apply the quadradic equation:\nx=(-b+/-(sqrt(b^2-4ac))/(2a)\n\nFor this equation, the equation is:\nx=(-(-1)+/-(sqrt((-1^2)-4*2*(-1))/(2*(2))\n\nSimplified it is:\nx=(1+/-sqrt(1+8))/(4)\nx=(1+3)/4 or x=(1-3)/4\n\nSo, x=1 or x=-1/2
926880
(1+1)-2*3/2^2+2\n\nWithout useing Order of Operations: 2\n\nWith Order of Operations: 2.5
267409
(-b + or - the square root of (b squared - 4AC)) divided by 2A