How do you rationalize the denominator and simplify #(x - 42) / (sqrtx+7) - 7#?

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Answer 1 · 2017-07-03T18:26:22

See a solution process below:

To rationalize the denominator multiply the expression by #(7 - sqrt(x))/(7 - sqrt(x))#

#(7 - sqrt(x))/(7 - sqrt(x)) xx (x - 42)/(sqrt(x) + 7) - 7 =>#

#(7x - 294 - xsqrt(x) + 42sqrt(x))/(7sqrt(x) + 49 - x - 7sqrt(x)) - 7 =>#

#(7x - 294 - xsqrt(x) + 42sqrt(x))/(49 - x) - 7#

To subtract the #7# we need to put it over a common denominator:

#(7x - 294 - xsqrt(x) + 42sqrt(x))/(49 - x) - (7 xx (49 - x)/(49 - x)) =>#

#(7x - 294 - xsqrt(x) + 42sqrt(x))/(49 - x) - (343 - 7x)/(49 - x) =>#

#(7x - 294 - xsqrt(x) + 42sqrt(x) - 343 + 7x)/(49 - x) =>#

#(7x + 7x - xsqrt(x) + 42sqrt(x) - 294 - 343)/(49 - x) =>#

#(14x + (42 - x)sqrt(x) - 637)/(49 - x)#