SOLUTION: Rationalize the function and give its domain in interval notation: F(x) = {{{ (2+sqrt(x-3))/(5+sqrt(x-3)) }}} So far for my work: I first, multiplied by the conjugate; {{{ 5-sq

Algebra ->  Rational-functions -> SOLUTION: Rationalize the function and give its domain in interval notation: F(x) = {{{ (2+sqrt(x-3))/(5+sqrt(x-3)) }}} So far for my work: I first, multiplied by the conjugate; {{{ 5-sq      Log On


   

Question 1121807: Rationalize the function and give its domain in interval notation:
F(x) = +%282%2Bsqrt%28x-3%29%29%2F%285%2Bsqrt%28x-3%29%29+
So far for my work: I first, multiplied by the conjugate; +5-sqrt%28x-3%29+
I ended up with +%28x%2B13%2B3sqrt%28x-3%29%29+%2F+%28x%2B28%29+
I don’t think that’s right and I’m having trouble knowing what the correct answer is as my text does not have this answer solution to check.
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
%282%2Bsqrt%28x-3%29%29%2F%285%2Bsqrt%28x-3%29%29

Yes, multiply the entire expression by 1=%285-sqrt%28x-3%29%29%2F%285-sqrt%28x-3%29%29, and simplify from that multiplication.

Result of that after simplification should be
F%28x%29=%28-x%2B13%2B3sqrt%28x-3%29%29%2F%2828-x%29

Or if multiply again by 1=%28-1%29%2F%28-1%29 you could have as
highlight%28F%28x%29=%28x-13-3sqrt%28x-3%29%29%2F%28x-28%29%29

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The radical forces the restriction x%3E=3 and the denominator cannot be allowed 0, so forces the restriction x%3C%3E28.