How do you rationalize the denominator #2/(5-sqrt3)#?

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Answer 1 · 2015-06-03T08:50:49

#2/(5-sqrt3)#

We can ratioinalise the denominator by multiplying the numerator and denominator of the expression by the conjugate of #5-sqrt3#

The conjugate of #5-sqrt3 = color(red)(5+sqrt3#

#=(2xxcolor(red)((5+sqrt3)))/((5-sqrt3)xxcolor(red)((5+sqrt3))#

We know that #color(blue)((a-b)(a+b)=a^2 - b^2 #
so ,#(5-sqrt3)xx(5+sqrt3) = 5^2- sqrt3^2 = color(blue)(25 - 3 = 22#

the expression becomes:
#(cancel2xxcolor(red)((5+sqrt3)))/cancel22#

# = (5+ sqrt3)/ 11#