How do you rationalize the denominator and simplify #5 / ( sqrt 14 - 2 )#?

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Answer 1 · 2016-04-08T06:44:22

#(sqrt14 + 2)/2 = sqrt14/2 + 1#

By rationalizing a fraction, we mean removing any irrational values from the denominator, without changing the fraction.

Here, we have to remove #sqrt14# from the denominator, without changing the value of the expression.

We know that #a^2-b^2 =(a+b)(a-b)#

We have #sqrt14-2# as the denominator. To remove the square root, we must multiply the denominator by #sqrt14+2#

Dividing and multiplying a fraction by the same number does not change the fraction.

#5/(sqrt14 -2) xx (sqrt14+2)/(sqrt14 + 2)#

#(5 xx (sqrt14+2))/((sqrt14 -2)(sqrt14 + 2))#

#(5sqrt14 + 10)/((sqrt14)^2 - 2^2)#

#(5sqrt14 + 10)/(14 -4) =(5sqrt14 + 10)/10#

5 is the common factor of the numerator.

#(5(sqrt14 + 2))/10#

Cancel 5 from the numerator and denominator.

#(sqrt14 + 2)/2 = sqrt14/2 + 1#