What is the coefficient of x^2 in the expansion of (5+2x)^0.5?

  1. We the answer in question in the form A(1+B)0.5 and we can do that by factorising out the 5. This gives 50.5(1+0.4x)0.5. 2. Identify the formula we are going to use and put in the numbers from this equation to get 50.5(1 + 0.50.4x + (0.5-0.50.4/2!)(0.4)2x2 + ... 3. At this point we have the x2 term so no further working is needed other than to multiply out the coeffiicient in front of the term to give -0.0250.5 = - 0.0447 (to 3SF)
AA

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