how do you solve or simplyfy 2x^2+4/2x?
117786
Well, there isn't really anything to "solve", since it's not an equation.  If you want to simplify, it depends what your original terms are.  Is the question\n\na)  2x^2 + 4/2x  ?\nIf so, then the only simplification is 2x^2 + 2/x.\n\nb) (2x^2 + 4) / 2x ?\nIn this case, the simplification is (x^2 + 2) / x.\n\nI hope this answers your question!  Good luck.
528350
(5x + 4)(6x + 4)\n30x^2 + 20x + 24x + 16\n30x^2 + 44x + 16\nANS : 16
1054159
First, get rid of the brackets (multiply the terms inside the brackets by -1):\n\n7-2x^2 +2x -2 -2x -5 +2x^2\n\nCollect like terms:\n\n2x^2 -2x^2 +2x -2x +7-2-5 = 0\n\nIt's easy to see the sum is zero.
920734
2xy(-15x^2y^4)+8x^2)-5/-4xy^3\n(-1/2)[-15x^2y^4+8x^2-5)/y^2\n(-1/2y^2)[-15x^2y^4+8x^2-5]
985516
(8x^4 + x^2 + 9)/x^4\n= 8x^4/x^4 + x^2 / x^4 + 9/x^4\n= 8 + 1/x^2 + 9/x^4 \nor 8 + x^-2 + 9x^-4
1160760
what you have is something like\na(b+c)\n\nthe solution is \nab+ac\n\n(x^3)(2x^-2)+(x^3)(4x)\nnow to multiply x^a * x^b you get x^(a+b)\n\n2x^((3+(-2))+4x^(3+1)\n\n2x+4x^4
764068
2x - 2 = x^2 - 4x + 3\nSet one side equal to 0\n0 = x^2 - 6x + 5\nFactor\n0 = (x - 5)(x - 1)\nSet each factor equal to 0\n0 = x - 5              0 = x - 1\nSolve\n5 = x                  1 = x
1053464
if you want to simplify it \n\nit is 3x - 4x + 4 = -x + 4
227193
∫[(2x^4) - (4/x^5)]dx=\n\n∫(2x^4)dx - ∫(4/x^5)dx=\n\n2∫x^4dx - 4∫[x^(-5)]dx=\n\n2*[x^5/5] - 4*[x^-4/-4]=\n\n(2/5)x^5 + x^-4 + C=\n\n(2/5)x^5 + 1/x^4 + C
926616
2x+3<6x-1\n2x-6x<-1-3 \n-4x<-4\nx>1
520950
Is this 2x^2?\n\nIf so, (-b +/- square root b^2 - 4ac)/2a  the formula\n(-12 +/- sqrt 12^2 - 4(2)(-8))/ 2(2)  (Plug in a, b, c)\n(-12 +/- sqrt 144 + 64)/4 (Simplify square root)\n(-12 +/- sqrt 208)/4  (Simplify square root)\n(-12 +/- 4 sqrt 13)/4  (Simplify square root)\n(-3 +/- sqrt 13)   (Divide all terms by 4)\n\nAnswer:  -3 + square root 13 and -3 - square root 13
769323
6 - 12x\n-or-\n-12x + 6\n\n-or-\n\n6(1-2x)\n\n...although that depends on whether you want the simplest expanded form or the simplest factored form.
303479
ok. \n\nThe trick is this. SQUARE THE EQUATION.\nNote that : [{2 + (3^1/2)}^1/2]^x =  [{2 + (3^1/2)}^(x/2)] \n\nWe get\n\n[ {2 + (3^1/2)}^1/2]^x + [{2 - (3^1/2)}^1/2]^x = 4\n\n{2 + (3^1/2)}^x \n+ 2 [{2 + (3^1/2)}^(x/2)]*[{2 - (3^1/2)}^(x/2)]  \n+ {2 - (3^1/2)}^x = 16\n\nand [{2 + (3^1/2)}^(x/2)]*[{2 - (3^1/2)}^(x/2)]  equals to\n [ 4 - 3 ] ^(x/2) = 1.\n\nThus, we get,\n{2 + (3^1/2)}^x + 2 + {2 - (3^1/2)}^x = 16\n{2 + (3^1/2)}^x + {2 - (3^1/2)}^x = 14  ...(1)\n\nNow note that \n[{2 + (3^1/2)}^(x)]*[{2 - (3^1/2)}^(x)]  = [4 - 3]^x = 1\nmeans that,\n{2 - (3^1/2)}^x = 1 / [{2 + (3^1/2)}^(x)]\nSubstituiting this into (1), we get,\n{2 + (3^1/2)}^x + 1 / [{2 + (3^1/2)}^(x)]  = 14  \n\nLet y = [{2 + (3^1/2)}^(x)] \nThe equation above becomes,\n\n y + 1 / y = 14\n y^2 - 14y + 1 = 0 \n\nA quadratic equation in y.\nWhich gives y = [14 ± sqrt(192)]/2 \n              \nSince y = [{2 + (3^1/2)}^(x)], Thus,\n\n [{2 + (3^1/2)}^(x)] = [14 ± sqrt(192)]/2 \n                     = [14 ± 8sqrt(3)]/2\n                     =  7 ± 4*3(1/2)\nSo,\n    x = ln [ 7 ± 4*3^(1/2)] / ln [2 + (3^1/2)]\n      =   ± 2\n\nAlternately, \nOne can use the identity,\n{2 + (3^1/2)}^(2) = 7 + 4*3(1/2)\n\n{2 + (3^1/2)}^(-2) = 7 - 4*3(1/2)\n\nTo get x = ± 2 again.\n\nDone.\n\n \nThe below is ADDITIONAL information solving a slightly different problem though similar to the one you posted.\n\nSay we have a similar problem but x is a positive real.\nWe want to find solutions for \n [ 2 + {(3^1/2)}^1/2]^x + [2 - {(3^1/2)}^1/2]^x = 4\n\nBy observation, x = 1 is a solution since, \n\n[ {2 + (3^1/2)}^1/2] + [{2 - (3^1/2)}^1/2] = 4\n\nHow to show they arent any other solutions?\n\n( x > 0 )\n\nLet f(x) =[ {2 + (3^1/2)}^1/2]^x + [{2 - (3^1/2)}^1/2]^x - 4\nSo, our equation amounts for solutions for f(x) = 0.\nWe show that f(x) is increasing for x>0 and we are done.\nNote that when we show f(x) increasing, it means that \nf(x)=0 only at one value because for any x > y, increasing means f(x)> f(y).\nSince f(1) = 0, Thus,\nf(y) > f(1) = 0 for y > 1  and\nf(y) < f(1) = 0 for  0 < y < 1   \n\nAll that remains to be shown is that f(x) is increasing.\nSimplest way, is to use basic calculus. \nIts necessary and sufficent to show f'(x)>0 for x >0.\nUsing the fact for constants any constant m, if y = m^x,\nThen dy/dx = (m^x)(ln m) \n\nf'(x)=  [{2 + (3^1/2)}^1/2]^x *ln[{2 + (3^1/2)}^1/2]\n      + [{2 - (3^1/2)}^1/2]^x *ln[{2 - (3^1/2)}^1/2] \n\nNow, note that\n[{2 + (3^1/2)}^1/2] = 3.316074013\n[{2 - (3^1/2)}^1/2] = 0.683925987\nln[{2 + (3^1/2)}^1/2] = 1.198781557\nln[{2 - (3^1/2)}^1/2] = - 0.379905573\n\nSo,\nf'(x) = (3.316074013^x)*(1.198781557) \n        - (0.683925987^x)*(0.379905573)\n\nSo to show f'(x) >0, its equivalent to showing,\n\n(3.316074013^x)*(1.198781557) >(0.683925987^x)*(0.379905573)\n              (4.848586069)^x > (0.316909757)\n           x ln (4.848586069) > ln (0.316909757)\n                 1.57868713 x > -1.149138224\n                            x > - 0.727907513 \n\nwhich is true for all x > 0.\nThus, for x>0, the only solution to our modified\nproblem is x=1.\n\nAnd there is no "chance" in mathematics.\nIt must be reasoned out.\n\nRegards
983225
remember the rule nx^n-1\n2x^2-1\n \nSo in this case it is 2x
859187
I'll do one for u.\nIn the first,\nMultiplying both sides by x*,we get\nx+3x+2x*=7x*\n4x+2x*=7x*\nMoving 4x and 2x* to the R.H.S.\n5x*-4x+0=0\nThus 5x*-4x=0\nAttempt the others in the same manner.
19885
(5) x (2x^2-3x+5) I think
681236
Multiply each term in the first () by each term in the second ():\n\n2x times 2x = 4x^2\n2x times 1 = 2x\n-3 times 2x = -6x\n-3 times 1 = -3\n\nThen combine 2x and -6x\nAnswer: 4x^2 - 4x - 3