SOLUTION: a = 4 root6/root2 +root3 then the value of a+2*root2/a-2*root2 + a+2*root3/a+2*root3

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Question 593982: a = 4 root6/root2 +root3 then the value of a+2*root2/a-2*root2 + a+2*root3/a+2*root3

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
  • This is not a problem related to logarithms. You're more likely to get help if you post it in the right category, "Square root, cubic root, N-th root" or "Radicals -- complicated equations involving roots" would be best for this problem.
  • Do not use "root". There are an infinite number of types of roots. If it is a square root, use "square root" or "sqrt". If it is a cube root, use "cube root" or 3rd root, etc. Tutors are more like to help when the problems are expressed clearly.
  • If a numerator or denominator has more than one term, enclose it in parentheses so the tutors can tell what the numerators and denominators are. Again, tutors are more like to help when the problems are expressed clearly.
    Assuming "root" means square root the expressions you posted, without the parentheses, mean:
    a+=+4+sqrt%286%29%2Fsqrt%282%29+%2B+sqrt%283%29
    and
    a%2B2%2Asqrt%282%29%2Fa-2%2Asqrt%282%29+%2B+a%2B2%2Asqrt%283%29%2Fa%2B2%2Asqrt%283%29
    I doubt very much that either of these are correct.