SOLUTION: f(√(11)+√(15)) f(x) = x^2

Algebra ->  Radicals -> SOLUTION: f(√(11)+√(15)) f(x) = x^2      Log On


   

Question 612279: f(√(11)+√(15))
f(x) = x^2
Found 2 solutions by lynnlo, jsmallt9:
Answer by lynnlo(4176) About Me  (Show Source):
You can put this solution on YOUR website!
is this 1, or 2 problems

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Function problems will be a lot easier for you once you learn that the "x" in f(x) is just a placeholder. It represents whatever happens to be the input to the function. And the x%5E2 part of f%28x%29+=+x%5E2 tells use what the function does to its input, x. It squares it. No matter what the input is to this function f, it will square it. If the input is x, f will square it. If the input is 5, f will square it. If the input is q, f will square it. If the input is a-4, f will square it.

If the input is sqrt%2811%29+%2B+sqrt%2815%29, as in f%28sqrt%2811%29+%2B+sqrt%2815%29%29, f will square it! So:
f%28sqrt%2811%29+%2B+sqrt%2815%29%29+=+%28sqrt%2811%29%2Bsqrt%2815%29%29%5E2

All that is left is to simplify. Exponents do not distribute! So it is not just 11 + 15. To square sqrt%2811%29%2Bsqrt%2815%29 correctly, use FOIL on %28sqrt%2811%29%2Bsqrt%2815%29%29%28sqrt%2811%29%2Bsqrt%2815%29%29 or use the %28a%2Bb%29%5E2+=+a%5E2+%2B2ab%2Bb%5E2 pattern. Personally, I prefer using the pattern:
f